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Daksh Quant Question Bank "May Edition" By Shantanu Shukla
Direction (1-5): Pie Chart (I) shows percentage distribution
of total number of students in five different schools and pie
chart (II) shows percentage distribution of total difference
between number of boys and girls in these five schools. Read
the data carefully and answer the questions.
Q1. When one student from each A & C chosen as class
monitor, then find the probability of both class monitors being
boys (boys > girls, for both the schools)?
(a)
(c) 84
(d)
(e)
A
20%
B
15%
D
25%
C
2
(a) 3
8
15
4
3
4
(e) 15
E
Total difference between boys
and girls = 50
Q3. Number of boys in school F are 5 less than that of in E and
number of girls in F are same as E. Find probability of selecting
1 boy & 1 girl from school – F. (Boys > girls, in school E)?
40
(a) 87
(b)
A
20%
(c)
14
29
43
87
2
(d) 3
(e)
B
20%
D
20%
A
2
56
(d) 5
D
E
30%
72
25
(c) 7
C
22.5%
B
25
Q2. If five students from each B & D are chosen for
representing their respective school in a debate competition,
then find the maximum possible probability of girls in
remaining students?
(b)
A
72
5
(b) 9
25
Total students = 200
E
17.5%
35
C
D
E
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Q4. Ratio of number of boys to girls left school A is 1 : 2 and
when two students chosen from remaining students of school
11
A, then probability of both being boys or both being girl is 18 .
Find difference between number of boys and girls who left the
school A (Boys > girls, in school A)?
(a) 6
(b) Can’t be determined
(c) 1
(d) 4
(e) 2
C
10%
B
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Q5. Ritu invested her total saving in three different FD
schemes A, B and C in the ratio of 5 : 4 : 6 on CI for two years
at the rate of 10%, 15% and 20% respectively.If interest is
calculated annually and interest from scheme B is Rs. 744
more than interest from scheme A then, find difference
between interest received from scheme C and scheme B by
Ritu?
(a) Rs. 4185
(b) Rs. 4175
(c) Rs. 3840
(d) Rs. 4580
(e) Rs. 3250
Direction (6–10): Bar graph given below shows percentage
of students passed out of total students in four different
schools and percentage of girls passed out of total passed
students in these four different schools in annual exam. Read
the data carefully and answer the questions.
70
60
50
Q7. Total boys failed from school Q is 60% of total students
failed from that school and total boys passed from same
school is 1440. Total failed boys from R is 192 more than total
female passed from same school. Find difference between
total girls failed from Q & R, if boys failed from R is 42% less
than total students failed from R.
(a) 272
(b) 242
(c) 252
(d) 240
(e) 262
Q8. If total number of girls passed from S is 1125 and total
1
failed boys are 133 % more than that of total failed girls from
3
same school, then find difference between number of failed
boys and failed girls from school S?
(a) 1100
(b) 1000
(c) 1110
(d) 900
(e) 1200
Q9. If difference between passed boys and passed girls from
P is 600 and total boys passed from S is 1350 , then find total
failed students from S is what percent more than total
students participated from school P ?
(a) 15%
(b) 10%
(c) 12%
(d) 5%
(e) 20%
40
30
20
10
0
P
Q
% of students passed
R
S
% of passed girls
Q6. Total boys passed from P is 900 and total girls failed from
Q is 640. If total girls failed from Q is 36% less than of total
students failed from Q, then find ratio of total students
participated in exam from Q to that of from P?
(a) 5 : 9
(b) 5 : 7
(c) 3 : 5
(d) 5 : 6
(e) 4 : 7
Q10. Total students participated in exam from school R is
50% more than that of from P and difference between passed
boys from both the school is 2220, then find average number
of girls passed from both the schools?
(a) 1100
(b) 1000
(c) 1390
(d) 900
(e) 1200
Directions (11-15): Line graph shows usual average speed of
five cars A, B, C, D and E. Speed is given in meter/minute.
Answer the question according to given data.
2000
1800
Speed (meters/minutes)
1750
1500
1500
1200
1250
1000
750
500
250
0
A
3
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900
750
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B
C
Cars
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E
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Q11. Car D started from Lucknow to Delhi at its usual speed
for first half distance, but after that due to some problem in
engine car travel at 4/6 of its usual speed. If car completed
whole journey in 10 hr, then find the total distance between
Lucknow and Delhi?
(a) 440 km
(b) 432 km
(c) 442 km
(d) 450 km
(e) 452 km
Q12. Car C starts from Pune and at the same time Car A starts
from Mumbai towards each other, and at the time both meet
one car has traveled 180 km more than other car. Find the
distance between Pune and Mumbai?
(a) 540 km
(b) 520 km
(c) 500 km
(d) 520 km
(e) 640 km
Direction (16-17): In the following wrong number series a
term is wrong as per pattern followed in the series. You have
to find that wrong term and term which should replace that
wrong term is your answer. (You have to mark correct term
as per series not the wrong term mentioned in series)
Q16. 32, 48, 80, 112, 176, 208, 272
(a) 36
(b) 116
(c) 206
(d) 276
(e) Series is right no need to replace any number
Q17. 12, 70, 348, 1390, 4168, 8332, 8332
(a) None of the given option will replace the wrong number
(b) 8334
(c) 4172
(d) 1392
(e) 350
Q13. Rajeev go to his village from the city by car B and return
by car C. If his total travelling time is of 11 hours, then find the
distance between city and his village?
(a) 500 km
(b) 510 km
(c) 520 km
(d) 540 km
(e) 1080 km
Direction (18-20): What will come in the place of (A), (B) &
(C) in second (II) number series. (II) number series follows
the same pattern as pattern of first (I) number series:
Q14. Car E travels at its usual speed between city X and Y and
take 480 minutes to complete total distance. But at the time of
returning car E decreases its speed by 12 km/hr. Then find
time taken by car E (in minutes) returning from city Y to X?
(a) 526 minutes
(b) 530 minutes
(c) 576 minutes
(d) 550 minutes
(e) 612 minutes
(c) 18.4, 34.4, 52
(d) 18.4, 34.4, 56
(e) 19.4, 34.4, 52
Q15. The distance between Delhi and Gorakhpur is 762 km.
Car E starts at 4 pm from Delhi towards Gorakhpur at a given
speed. Another car C starts at 3.20 pm from Gorakhpur
towards Delhi at a given speed. How far from Delhi both cars
meet and at what time?
(a) 8:20 pm, 312 km
(b) 7:20 pm, 290 km
(c) 8:10 pm, 390 km
(d) 6:20 pm, 350 km
(e) 9.20 pm, 480 km
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Q18. Series I – 2.4, 4, 7.2, 12, 18.4, 26.4, 36
Series II – (A), 20, 23.2, 28, (B), 42.4, (C)
(a) 18.4, 34.4, 54
(b) 18.4, 36, 52
Q19. Series I – 123, 171, 227, 291, 363, 443, 531
Series II – (A), (B), 402, 486, (C), 678, 786
(a) 258, 326, 580
(b) 258, 326, 576
(c) 256, 326, 578
(d) 258, 324, 578
(e) 258, 326, 578
Q20. Series I – 20, 1300, 2100, 2660, 3100, 3480, 3830
Series II – (A), 1600, 2400, (B), 3400, 3780, (C)
(a) 320, 2960, 4130
(b) 280, 2840, 4250
(c) 320, 2960, 4180
(d) 280, 2840, 4200
(e) 320, 2960, 4050
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Directions (21-25) : An exam consists of 100 questions
among 3 sections i.e. A, B and C. Section C have 32 questions
and section A and B have equal number of questions. For each
correct answer a student is awarded 3 marks and 1 mark is
deducted for every wrong answer. Each section has 3
questions for which there is no negative marking on wrong
attempt and 0.50 marks are deducted for every unattempted
question.
Amar and Prem appeared for same exam, Amar attempted
total 75 questions, while Prem attempted 80% of total
questions attempted by Amar and 90% of his questions were
correct. Both of them attempted all the question which do not
have negative marking.
Q21. What is the smallest range of marks obtained by Amar,
if 80% of his attempted questions were correct?
(a) 165-174
(b) 165-174
(c) 152.5-161.5
(d) 165-180
(e) 155-165
Q22. If only 3 non-negative marking questions out of 9 were
wrong for Prem. Find the score of Prem.
(a) 122
(b) 102
(c) 159
(d) 139
(e) None of these
Q23. Prem attempted all the questions of section A, which
were all correct and 50% of questions section C in which 6
were incorrect carrying negative marking. Find the difference
between his score from section C and section B.
(a) 03
(b) 02
(c) 14
(d) 06
(e) None of these
Q24. All the non-negative marking questions were correct for
Amar and only non-negative marking questions were wrong
for Prem. What is the difference between their score, If 80%
of questions attempted by Amar were correct.
(a) 10.5
(b) 16.5
(c) 54.5
(d) 36.5
(e) 17.5
Q25. If Amar scored 108.5 and 5 of his non-negative marking
questions were correct. What is the total number of his
correct questions?
(a) 42
(b) 53
(c) 47
(d) 49
(e) None of these
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Directions (26-30): Study the graph and table given below
and answer the following questions. The line graph shows the
listed price per kg of various items in a wholesale store.
45
40
35
30
25
20
15
10
5
0
The table given below shows the amount of items bought by a
retailer from the wholesale store. The table also shows the
discount % offered by the wholesaler on the list price and
total cost incurred by the retailer.
Quantity
Discount
Total
Items
(in kgs)
(in %)
(in Rs.)
Rice
20
10
Wheat
30
675
Sugar
15
240
Pulses
18
30
Cashew
40
900
Almond
25
15
Q26. Calculate the profit earned by retailer on selling 20 kgs
of wheat purchased by him to a customer at a discount of 5%
on the listed price ?
(a) Rs. 25
(b) Rs. 45
(c) Rs. 75
(d) Rs. 50
(e) None of these
Q27. The retailer sold all the cashew bought by him to a
customer at a price 25% more than the listed price. Calculate
his overall profit percent.
(a) 33.33%
(b) 66.66%
(c) 55.55%
(d) 42.64%
(e) 77.77%
Q28. If 50% of the rice bought by the retailer got spoiled, then
at what price/kg must he sell the remaining amount of rice to
be at a situation of no loss-no gain ?
(a) Rs. 40
(b) Rs. 19
(c) Rs. 27
(d) Rs. 22
(e) None of these
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Q29. The retailer sold all the pulses he bought at a price that
is 30% more than the listed price and offered 2 kgs of Almond
free with it. Find overall profit% of the retailer in this bargain?
(approximate)
(a) 50%
(b) 40%
(c) 35%
(d) 61%
(e) 45%
Q30. The retailer mixed 6 kgs. of impurity (free of cost) with
all the sugar he had and sold the mixture at a discount which
is 25% less than that discount (in percentage) offered by the
wholesaler. Find the profit % on the sale of all of the amount
of this mixture?
(a) 52.50%
(b) 46.15%
(c) 48.75%
(d) 57.50%
(e) None of these
Direction (31–35): Table given below shows data regarding
number of people applied for loan under ‘PM Mudra Yojna’
from five different villages. Read the data carefully and
answer the questions.
Ratio
of
Percentag
Percentag
femal
e of people
e of male
e who
Numbe
who get
who get
do not
r of
loan out of
Village
loan out of
get
people
total
s
total
loan
applied number of
people
to
for loan
people
who get
femal
applied for
loan
e who
loan
get
loan
P
7200
66 3 %
2
65%
10 : 21
Q
8000
60%
75%
3:5
R
8800
9
81 11 %
82%
4:9
S
T
10000
9600
72 %
68.75%
76%
80%
16 : 27
8 : 11
Q31. What is the difference between number of males applied
for loan from village P and village T?
(a) 2600
(b) 2200
(c) 2400
(d) 3000
(e) 2000
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Q32. Total females who do not get loan from village S is what
percent more or less than total females who do not get loan
from village Q?
2
(a) 40 9 %
2
(b) 44 %
9
2
(c) 46 %
3
2
(d) 48 3 %
2
(e) 42 9 %
Q33. What is the ratio of total males who do not get loan from
village Q to total males who do not get loan from village S?
(a) 113:111
(b) 115: 111
(c) 64: 111
(d) 155: 111
(e) 111 : 115
Q34.How many females applied for loan from the village
where number of males who get loan are second highest
among all villages?
(a) 2772
(b) 2726
(c) 2752
(d) 2742
(e) 2732
Q35. What is the average of number of males who do not get
loan from village R and number of females who get loan from
village R?
(a) 1120
(b) 1140
(c) 1260
(d) 1200
(e) 1160
Direction (36-40): Find the wrong number in the following
number series given below :
Q36. 5, 86, 174, 276, 399, 558, 736
(a) 276
(b) 736
(c) 558
(d) 86
(e) 399
Q37. 9, 4.5, 6.5, 14, 57, 457, 7313
(a) 4.5
(b) 57
(c) 457
(d) 9
(e) 7313
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Q38. 1728, 998, 1511, 1167, 1384, 1260, 1323
(a) 998
(b) 1511
(c) 1323
(d) 1167
(e) 1260
Directions (44-45): Answer these questions based on the
information given below.
6 men complete a piece of work in 12 days. 8 women can
complete the same piece of work in 18 days. Whereas 18
children can complete the piece of work in 10 days. 4 men, 12
women and 20 children work together for 2 days, and then
only 36 men were to complete the remaining work in 𝑥 day.
Q39. 2.5, 60, 720, 4320, 12960, 19480, 14580
(a) 720
(b) 4320
(c) 12960
(d)19480
(e) 14580
Q44. A can do a piece of work in 10𝑥 days and B can do the
same work in 20𝑥 days. They do the work on alternative days
starting from A then in how many days A and B can complete
the work?
(a) 11 days
(b) 12 days
(c) 13 days
(d) 14 days
(e) None of these
Q40. 11.5, 34, 58, 85, 116.5, 154, 200
(a) 200
(b) 85
(c) 116.5
(d) 154
(e) 34
Q41. Working alone, A can complete a task in ‘a’ days and B in
‘b’ days. They take turns in doing the task with each working
2 days at a time. If A starts, they finish the task in exactly 10
days. If B starts, they take half a day more. How long does it
take to complete the task if they both work together?
1
(a) 5 days
3
1
(b) 5 7 days
5
(c) 5 9 days
5
(d) 5 11 days
(e) None of these
Q42. Bag A contains ‘P’ green and 18 yellow balls while bag B
contains ‘(P + 2)’ green balls and 22 more number of yellow
balls than that of in bag A. Probability of selecting a green ball
1
from bag A is 12 more than probability of selecting a green ball
from bag B. Find total number of balls in bag B. (P<50)
(a) 48
(b) 54
(c) 60
(d) 84
(e) 66
Q45. 56𝑥 soldiers can complete a piece of work in 24 days. In
how many days can 42 soldiers complete the same piece of
work?
(a) 32 days
(b) 24 days
(c) 16 days
(d) 48 days
(e) None of these
Directions (46-47): Three friends P, Q and R share an
apartment and share the rent equally. The monthly income of
R is 25% less than that of Q and Rs.8000 less than that of P.
Monthly expenditure of Q on food is Rs.1000 more than that
of P and is Rs.1000 less than that of R. After meeting the
expenses on rent and food, they save amounts in the ratio 6 :
7 : 4.
1
Q46. If Q saves 62 2 % of his total monthly income, then how
much percent does R save out of his monthly income?
13
(a) 47 21 %
12
(b) 48 21 %
5
(c) 45 21 %
11
(d) 49 %
21
(e) Cannot be determined
Q43. Ratio of speed of boat P and Q in still water is 2 : 3. If time
taken by boat ‘Q’ to cover a certain distance in upstream is
equal to that of ‘P’ to cover 60% of that distance in upstream
and difference of the distance covered by ‘Q’ and ‘P’ in 5hr in
downstream is 60km then find ratio of speed of boat ‘P’ in
downstream to ‘Q’ in upstream?
(a) 3 : 2
(b) 1 : 2
(c) 1 : 1
(d) 4 : 5
(e) 5 : 4
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Q47. If the total amount spent by all the three on food is
Rs.27000 and the monthly income of Q is Rs.6000 more than
that of P, then what is the monthly rent of the apartment?
(a) Rs.48000
(b) Rs.30000
(c) Rs.24000
(d) Rs.36000
(e) Cannot be determined
Q48. Selling price of two articles is same and shopkeeper
claim to get 20% profit on each article but mistakenly he
calculates one profit on selling price. If difference of profit he
gets on articles is 8 Rs. then find the selling price?
(a) Rs.200
(b) Rs.180
(c) Rs.160
(d) Rs.220
(e) Rs.240
Q52. If the number of girls in sections A is same as the number
of boys in section C, then what is the ratio of number of boys
in section A to the number of boys in section C?
(a) 2 : 3
(b) 3 : 4
(c) 3 : 2
(d) 4 : 3
(e) None of these
Directions (53-55): Given below are two pie charts. Pie chart
(1) shows the percentage distribution of milk in five vessels
out of the total milk in these five vessels. Pie chart (2) shows
the percentage distribution of water in same five vessels out
of total quantity of water in these five vessels.
(1) (2)
E
20%
Q49. Rs.15000 invested in two schemes each. First scheme
offers R% interest on C.I. and second scheme offer R% more
of what he gets in first at S.I. If after 2-year difference between
interests earned in both scheme is Rs. 600 then find the value
of R.
(a) 24%
(b) 10%
(c) 15%
(d) 20%
(e) None of these
A
15%
B
20%
D
10%
C
35%
Q50. Sunny and Satish shoot each other 4 and 3 out of every
6 shots respectively. They try to shoot each other in sequences
start with Sunny. Find the probability that Satish killed?
1
(a) 5
D
5%
E
15%
A
10%
B
25%
2
(b)5
7
(c)10
C
45%
3
(d)5
4
(e)5
Directions (51-52): There are three sections A, B and C in a
class. Every section has some boy and some girl students in it.
Probability of a girl being selected when one student is
2
4
selected randomly from section A is 5, that from section B is 9
5
and that from section C is .
9
Q51. If the ratio of total number of students in sections A, B
and C is 10 : 12 : 9, then what is the probability of a girl being
selected when one student is selected randomly from the
students from all the three sections together?
14
(a) 31
(b)
(c)
Note: Ratio of total milk to total water in these five containers
is 2 : 1.
Q53. A shopkeeper pours the mixture of vessel A and B into
another vessel F. Vessel F contains water only which is equal
to 25% of water of vessel B. If shopkeeper professes to sell the
whole mixture at the cost price of pure milk and cost price for
shopkeeper is due to milk only, then find the percentage profit
of shopkeeper in selling whole mixture.
13
(a) 5814%
13
11
(b) 315%
23
13
(c) 54 %
31
43
(d) 53 %
13
15
13
(d) 93
(e) Cannot be determined
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20
(e) 5521%
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Q54. Mixture of vessel A and C are mixed into another vessel
M. If 62 liters of the mixture M is taken out and replaced with
17 L of water, the ratio of milk to water in M becomes 6 : 5.
Find the quantity of milk in vessel B.
(a) 60 L
(b) 20 L
(c) 40 L
(d) 45 L
(e) 50 L
Q55. All the contents of mixture from all vessels except C is
poured into bigger vessel and from vessel C, only 115 liters of
mixture is taken out and poured into bigger vessel, then ratio
of milk and water in bigger vessel becomes 9 : 4. Find the total
quantity of water in all five vessels.
(a) 550 L
(b) 500 L
(c) 600 L
(d) 650 L
(e) 700 L
Directions (56-60): The following questions are
accompanied by three statements (A), (B), and (C). You have
to determine which statement(s) is/are sufficient necessary
to answer the questions
Q56. A bag contains some balls of red, black and blue colors.
3 balls are drawn randomly from it. What is the probability
that the balls drawn are of three different colors?
A. Ratio of black and blue colored balls is 4 : 5.
B. Total number of red and black balls is 5 more than that of
blue colored balls.
C. Total number of balls in that bag is 35 and probability of
3
selecting blue colored balls is 7.
(a) A and either B or C
(b) Any two of them
(c) Only A and C together
(d) Question can't be answered even after using all the
information
(e) All statements are required
Q57. Find area of a rectangle if its perimeter is equal to the
perimeter of a square.
A. Length and breadth of the rectangle are in the ratio of 4 :3
and side of a square is half of the sum of length and breadth of
that rectangle.
B. Sum of the lengths of diagonals of the rectangle is 4m less
than thrice of the side of square.
C. Area of a square is 784 m2
(a) Only A and B together
(b) Only A and C together
(c) All the three together
(d) Any two of the three together
(e) Either A and B together or A and C together
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Q58. What is the value of a two digit number?
A. The sum of the number and its square is 25 times the
number itself.
B. The number obtained after interchanging the digits is
greater than the original number by 18.
C. The ratio of the value of the number and the sum of the
digits of that number is 4 : 1.
(a) Either A alone or B and C together
(b) All the three together
(c) Any one of the three
(d) All the three together are not sufficient
(e) Either A or B alone
Q59. What is the selling price of an article if the marked price
of that article is 60% above the cost price of that article?
A. The article is sold at 12% profit and there are two
1
successive discounts of 20% and 12 2 % on marked price.
B. Difference of marked price and the cost price of that article
is Rs 480.
C. The marked price is Rs.384 more than the selling price of
that article.
(a) Only B and C together
(b) Any two of three are sufficient
(c) Either A and B together or A and C together
(d) Only A and C together
(e) All the three together
Q60. Three friends A, B and C invested in a scheme. Find the
profit share of A?
A. Investment of A, B and C is in the ratio 9: 12: 10 and share
of C in profit is Rs.1800
B. A and B invested Rs.8100 and Rs.10800 for 10 months
respectively and profit share of B is 100% of the profit share
of C.
C. A invested Rs 900 more than that of C and investment
period of B is 10 months.
(a) A and B together only
(b) Either A and B or A and C
(c) Any two of them
(d) Either B alone or A and C together
(e) Either A and C or B and C
Directions (61-65): The following questions are
accompanied by three statements A, B and C. You have to
determine which statement(s) is/are necessary/sufficient to
answer the question.
Q61. A shopkeeper sells articles at a certain profit. Find out
the amount of profit.
A. Ratio of the selling price to the cost price of the articles is 5
: 4.
B. If the cost price increases by Rs.500, and selling price
8
remains the same, the profit percentage is decrease by 139
percentage points.
C. If the marked price is kept at Rs.1000 above the cost price
and a discount of 15% is given, then the profit percentage is
3
decreased by 18 4 percentage points.
(a) Only A and B together
(b) A and either B or C
(c) Only A and C together
(d) All statements are required
(e) None of these
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Q62. Rinku borrowed an amount of Rs.5000 each from Milan
and Rahul. What is the rate of interest?
A. Rinku returned the amount of Rs.5400 after due date to
Milan.
B. Rinku returned Rs.5900 to Rahul after due date.
C. Rinku returned the money to Milan by SI, whereas to Rahul
by compound interest.
(a) Only A and B together are sufficient
(b) Only B and C together are sufficient
(c) A, B and C together are necessary
(d) Either A and B together or B and C together are sufficient
(e) A, B and C even together are not sufficient
Q63. What is the speed of boat in still water?
A. The boat can cover 45 km downstream distance in 3 hours.
B. Speed of the stream is one-fourth the speed of boat in still
water.
C. The boat can cover 36 km upstream distance in 4 hours.
(a) Only (A) and (C) together
(b) All the three together
(c) Only (A) and (B) together
(d) Questions can’t be answered even after using all the
information
(e) Any two of the three together
Q64. A train crosses another train in 10 sec. Find out the
lengths of the trains.
A. Ratio between the lengths the of second and first train is 4
: 5.
B. Ratio between the speed of first and second trains is 1 : 2.
C. The speed of first train is 36 km/hr.
(a) Only A and B together
(b) Only B and C together
(c) Only A and C together
(d) Questions can’t be answered even after using all the
information
(e) None of these
Q65. Find the height of an equilateral triangle.
A. Perimeter of the triangle is equal to the perimeter of the
rectangle whose length and breadth are in the ratio of 5 :3.
B. Perimeter of the square is 48 m, which is twice the
perimeter of the triangle.
C. Area of the triangle is 16√3m2.
(a) Any two of them
(b) Any of them
(c) Only C
(d) Either B or C alone
(e) A and either B or C
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Directions (66-70): In the given questions, two quantities
are given, one as Quantity I and another as Quantity II. You
have to determine relationship between two quantities and
choose the appropriate option
Q66. In a two-digit number, digit at unit place exceeds, the
digit in its tens place by 2 and the product of the required
number with the sum of its digit is equal to 144.
Quantity I: Value of two digit number
Quantity II: 26
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q67. Quantity I : Days after which A and B meet. A and B set
out to meet each other from two places 165 km apart. A
travels 15 km the first day, 14 km second day, 13 km the third
day and so on, B travels 10 km the first, 12 km the second day,
14 km the third day and so on.
Quantity II: Number of days required to complete the whole
work if A, B and C can complete a piece of work in 10, 12 and
15 days respectively. A left the work 5 days before the work
was completed and B left 2 days after A had left.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q68. Quantity I: Present age of Randy, if 10 years are
subtracted from the present age of Randy, then you would get
twelve times of the present age of his grandson Sandy and
Sandy is 19 years younger to Sundar whose age is 24.
Quantity II: Average age of the remaining persons in the
group if average age of group of 14 persons is 27 years and 9
months. Two persons, each 42 years old, left the group.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q69. Quantity I: Percentage profit earned by the shopkeeper
if at the time of selling and purchasing he uses weights 10%
less and 20% more per kilogram respectively and professes to
all goods at 5% profit.
Quantity II: ‘x’ ; A book was sold for a certain sum and there
was a loss of 20%. Had it been sold for Rs 12 more, there
would have been a gain of 30%. ‘x’ would be value of profit
percent if the book were sold for Rs 4.8 more than what it was
sold for.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
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Q70. A group consist of 4 couples in which each of the 4
persons have one
Quantity I : Number of ways in which they could be arranged
in a straight line such that the men and women occupy
alternate positions
Quantity II: Eight times the number of ways in which they be
seated around circular table such that men and women
occupy alternate position.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Directions (71-74): The following questions are
accompanied by two statements I and II. You have to
determine which statements(s) is/are sufficient/necessary to
answer the questions.
Q71. Find the sum of present age of X & Z?
I. Four years ago, average age of X, Y and Z was 16 years while
age of X, Y and Z is in arithmetic progression.
II. Average of present age of X and Z is same as present age of
Y while Y is four years younger than Z.
(a) Statement I alone is sufficient to answer the question but
statement II alone is not sufficient to answer the question.
(b) Statement II alone is sufficient to answer the question but
statement I alone is not sufficient to answer the question.
(c) Both the statements taken together are necessary to
answer the questions, but neither of the statements alone is
sufficient to answer the question.
(d) Either statement I or statement II by itself is sufficient to
answer the question.
(e) Statements I and II taken together are not sufficient to
answer the question.
Q72. Is Rahul meet Prabhat at half of the distance on a circular
track (48km) if they both started to move at same time in
same direction?
I. Speed of Prabhat is 24km/hr more than speed of Rahul.
II. Prabhat’s speed is 200% more than that of Rahul
(a) Statement I alone is sufficient to answer the question but
statement II alone is not sufficient to answer the question.
(b) Either statement I or statement II by itself is sufficient to
answer the question.
(c) Both the statements taken together are necessary to
answer the questions, but neither of the statements alone is
sufficient to answer the question.
(d) Statement II alone is sufficient to answer the question but
statement I alone is not sufficient to answer the question.
(e) Statements I and II taken together are not sufficient to
answer the question.
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Q73. Find number of yellow balls in a box which contains 3
black balls and 7 red and yellow balls.
2
I. Probability of selecting two yellow balls from the box is 15
1
II. Probability of selecting one red and black ball is 5.
(a) Statement I alone is sufficient to answer the question but
statement II alone is not sufficient to answer the question.
(b) Statement II alone is sufficient to answer the question but
statement I alone is not sufficient to answer the question.
(c) Both the statements taken together are necessary to
answer the questions, but neither of the statements alone is
sufficient to answer the question.
(d) Either statement I or statement II by itself is sufficient to
answer the question.
(e) Statements I and II taken together are not sufficient to
answer the question.
Q74. Find number of pens sold by retailer if he earns total
37.5% profit on selling some pens and some pencils.
I. Ratio between pen sold to pencil sold is 2 : 3 while ratio
between cost price of pen to pencil is 3 : 2.
II. On selling a pen and a pencil, he earns 25% and 50% profit
respectively.
(a) Both the statements taken together are necessary to
answer the questions, but neither of the statements alone is
sufficient to answer the question.
(b) Statements I and II taken together are not sufficient to
answer the question.
(c) Statement II alone is sufficient to answer the question but
statement I alone is not sufficient to answer the question.
(d) Statement I alone is sufficient to answer the question but
statement II alone is not sufficient to answer the question.
(e) Either statement I or statement II by itself is sufficient to
answer the question.
Directions (75-76): In the given questions, two quantities
are given, one as Quantity I and another as Quantity II. You
have to determine relationship between two quantities and
choose the appropriate option.
Q75. There are three vessel A, B and C. Vessel A contains
mixture of milk and water in ratio 3 : 1. Vessel B contains 20
litres of pure water while vessel C contains 30 litres of pure
milk. Half of the content of vessel A is first poured into vessel
B. Then content of vessel B is poured into vessel C and finally
contents of vessel C is poured into vessel A. The final ratio of
milk and water in vessel A is 9 : 4.
Quantity I: Initial quantity of mixture in vessel A.
Quantity II: 80 litres.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
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Q76. 20 men can complete a work in 12 days. 5 women are as
efficient as 3 men. 4 men and 10 women started working and
they already worked for 8 days.
Quantity I: Additional number of women required to
complete the remaining work in 10 days.
Quantity II: Additional number of men required to complete
the remaining work in either 8 or less than 8 days.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q77. Quantity I — Time taken by A to complete a work alone
if A can complete a work in 5 more days than B while A does
the same work in 9 more days than C. If A and B can complete
the whole work in same time as time taken by C alone to do
the whole work.
Quantity II — Time taken by 8 men and 14 women to reap
7
𝑝𝑎𝑟𝑡 of 360 hectare land by working 7 hrs per day if 6 men
12
5
and 10 women can reap 12 part of the land in 15 days by
working 6 hrs per day. It is also given that work of 2 men is
equal to that of 3 women.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q78. Quantity I — Difference between the speeds of P and Q
if 2 places A and B are 60 km apart. P and Q start from A at
same time & meet 1st time at a place 12 km from B & they
reach A after immediate return from B. The speed of slower
person is 48 km/hr.
Quantity II — Average speed of train if a distance of 600 km
is to be covered in 2 parts. In 1st phase 120 km is travelled by
train and rest by car and it took total of 8 hrs, but if 200 km is
covered by train and rest by car it takes 20 min more.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q79. Quantity I — Cost price of motor bike if a man promises
to pay the price of the motorbike in 3 equal annual
installments of 10,800 Rs. at the compound interest rate of
20% per annum.
Quantity II — 240% of the value of each installment if a man
borrowed a sum of Rs. 25220 from a bank and promise to pay
the amount in 3 equal installments at the compound interest
rate of 5% per annum.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
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Q80. Quantity I — Value of 𝑥, if ABCD is a rectangle and AB=
10 unit, AD= 6 unit.
Quantity II — value of 𝑦, if volume of the cone is 16𝜋 unit3
Radius = 4 unit
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Q81. Sandeep bought some chilli powder containing five
percent impurities in the form of raw dust. He then mixed
pure chilli powder to 2 kg of that impure chilli in such a way
that the proportion of impurity now became 4% of total
mixture. At what percent should he mark-up the price of this
mixture to have an overall profit of 30% ?
(a) 25%
(b) 35%
(c) 20%
(d) 32%
(e) None of these
Q82. In an examination, there are 100 questions. For each
correct answer 4 marks will be awarded, for each wrong
answer 2 mark will be deducted for each unattempted
question 1 mark will be deducted. If a student scored 320
marks then what could be max number of questions
attempted wrong by him?
(a) 12
(b) 10
(c) 14
(d) 13
(e)None of these
Q83. A and B can do a piece of work in 12 days. B and C
together can do the same work in 16 days while B alone can
do it in 24 days. Two of them who have same efficiency work
on day one and then the third person does the work on day
two. If this process goes on till the completion of the work,
then find the total no. of days taken for the work to get
finished ?
3
(a) 15 days
4
3
(b) 17 days
4
3
(c) 16 days
4
3
(d) 18 days
4
(e) None of these
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Directions (84-85): Abhishek bought some chairs and tables
from a shopkeeper. The marked price of a chair and a table
were in the ratio 5 : 8. The shopkeeper gave discounts of 20%
and 25% on the chair & the table respectively. The ratio of
number of chairs and tables bought by Abhishek is 6 : 5.
Q84. If Abhishek sells each chair and table bought by him at
discounts of 25% and 20% respectively after marking up the
prices of both by 50% and gives one table free for every four
2
chairs bought by a customer and only 3 rd of the total chairs
are sold in bunch of four chairs, then what is the net profit
/loss % made by Abhishek after selling all of the items which
he bought from the shopkeeper?
2
(a) 6 3 %
1
(b) 3 %
3
1
(c) 2 %
2
1
(d) 4 4 %
(e) None of these
Q85. If the marked price of a table set by the shopkeeper was
Rs.300 more than that of a chair and the total expenditure
made by Abhishek in purchasing the chairs and table from the
shopkeeper was Rs.108000, then how many chairs were
purchased by Abhishek?
(a) 150
(b) 60
(c) 120
(d) 90
(e) None of these
Directions (86-87): A team of miners planned to mine 1800
tons of ore during a certain number of days. Due to technical
difficulties, in one-third of the planned number of days, the
team was able to achieve an output of 20 tons of ore less per
day than the planned output. To make up for this, the team
overachieved for the rest of the days by 20 tons. The end
result was that the team completed the task one day ahead of
time.
Q86. How many tons of ore did the team initially plan to mine
per day?
(a) 50
(b) 100
(c) 150
(d) 200
(e) 250
Q87. To complete the task two days before the planned date,
how many more tons of ore should the team mine per day
after mining the planned tons of ores till one-third of the
planned no. of days?
(a) 25
(b) 30
(c) 20
(d) 15
(e) None of these
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Direction (88-89)- In a school there are a total of 50 students
in class X, who are divided into three sections A, B and C. A and
B have an equal number of students. All the students of the
class wrote a test. The average marks obtained by the
students of sections A and B together is 52.5. The average
marks obtained by the students of sections A and C together
is 60. The average marks obtained by the students of sections
B and C together is 70. The average marks obtained by the
students of sections A, B and C together is 60.
Q88. How many students are there in section C?
(a) 10
(b) 20
(c) 15
(d) Cannot be determined
(e) None of these
Q89. How many students are there in all of the three sections?
(a) 40
(b) 30
(c) 45
(d) Cannot be determined
(e) None of these
Q90. Ratio of length and speed of two trains (smaller: longer)
are 7: 8 and 4: 3 respectively. Time taken by both trains to
cross each other while running in opposite directions is
6
12 7 𝑠𝑒𝑐 and time taken by smaller train to cross a platform
having length 100 m less than the smaller train is 16 sec. Find
time taken by longer train to cross that platform?
1
(a) 22 2 𝑠𝑒𝑐
1
(b) 23 3 𝑠𝑒𝑐
2
(c) 21 3 𝑠𝑒𝑐
(d) 20 sec
(e) None of the given options
Directions (91-93): The following questions are
accompanied by two statements (I) and (II). You have to
determine which statements(s) is/are sufficient/necessary to
answer the questions.
(a) Statement (I) alone is sufficient to answer the question but
statement (II) alone is not sufficient to answer the questions.
(b) Statement (II) alone is sufficient to answer the question
but statement (I) alone is not sufficient to answer the
question.
(c) Both the statements taken together are necessary to
answer the questions, but neither of the statements alone is
sufficient to answer the question.
(d) Either statement (I) or statement (II) by itself is sufficient
to answer the question.
(e) Statements (I) and (II) taken together are not sufficient to
answer the question.
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Q91. What is volume of cone?
I. Radius of cone is 3 cm less than side of square, whose area
is 576 cm2.
II. Height of cone is 7.5 cm more than radius of circle, which
circumference is 66 cm.
Q92. What is rate of interest?
I. A man invested an amount for three years on simple interest
and gets a total amount, which is 137.5% of invested amount.
II. Amir invested Rs. 9600 on simple interest and gets a total
amount of Rs.13200 after three years.
Q93. The ratio between length of two trains is 9 : 8. What will
be difference between lengths of both trains?
I. Speed of larger train and smaller train is 72 km/hr and 90
68
km/hr respectively. Both trains cross each other in
𝑠𝑒𝑐
9
running in opposite direction.
II. Speed of smaller train is 90 km/hr and it cross a pole in 6.4
sec.
Directions (94-96): In the given questions, two quantities
are given, one as ‘Quantity I’ and another as ‘Quantity II’. You
have to determine relationship between two quantities and
choose the appropriate option:
Q94. Quantity I – A bag contains five red balls, six green balls,
‘a’ yellow balls & ‘b’ blue balls. Probability of drawing one
1
yellow ball is , while probability of drawing one blue ball is
6
2
. If two balls are drawn from bag without replacing, then find
9
probability that one of them is red and other is yellow.
Quantity II – A bag contains dices only in three colors, eight
green color dice, ‘x’ blue color dice and ‘y’ yellow color dice.
7
Probability of drawing one blue dice is 20 , while probability of
1
drawing one yellow dice is 4. If two dices drawn at random
without replacement, then find probability that one of them is
blue and the other is green.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or no relation
Q95. Quantity I – A cylindrical vessel with radius and height
of 17.5 cm and 18 cm respectively is filled upto 80% of its
capacity with milk. If total milk from cylindrical vessel
transferred into 30 cuboidal vessels whose length and
breadth is 7 cm & 3 cm respectively. Find height of each
cuboidal vessel?
Quantity II – Breadth of a rectangle is 18 cm and ratio
between length of rectangle and side of square is 12 : 11. If
perimeter of square is 4 cm more than perimeter of rectangle.
Find side of square.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or no relation
14
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Q96. Quantity I – A boat takes double time in covering same
distance in upstream as compared to downstream, if boat
covers 96 km in downstream and 72 upstream in total 20
hours. Find time taken by boat to cover 240 km in
downstream.
Quantity II – Distance between point A and point B is 720 km.
1
A car covered 𝑟𝑑 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 with its usual speed and
3
remaining with 20% increased speed, if car takes total 10
hours 40 minutes to cover total distance, then find in what
time car will cover a distance of 1200 km with its usual speed.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or no relation
Q97. A bank offers SI of 10% on principal amount below
Rs.5000 otherwise 20%. A man invested Rs. A for 3 years.
What is the value of A?
(A) If he had submitted Rs. 3000 more, he will get an interest
of Rs. 900 more.
(B) If he had submitted Rs. 4000 more, he will get an interest
of Rs. 2400 more.
(C) Value of A is multiple of 500.
(a) Either A and B or B and C are sufficient to answer the
question
(b) Either A and B or A and C are sufficient to answer the
question
(c) Either A and C or B and C are sufficient to answer the
question
(d) A, B and C together are sufficient to answer the question
(e) All of the statements together are not sufficient to answer
the question.
Q98. A man has two items A and B. What is the selling price of
item B?
(A) The amount obtained after selling 4 of item A and 1 of item
B is Rs. 70. From this amount he either could buy 7 of item A
or 1 of item A with 4 of item B.
(B) Profit earned on selling 1 of item B is Rs. 3 and profit %
earned on selling 1 of item A is 30 %. Selling price and cost
price of item B are Rs. 5 greater than that of item A
respectively.
(C) Ratio of profit % earned on A and B is 3 : 2 and ratio of
their cost price is 2 : 3 respectively. All profit % , cost price
and selling price have integer value.
(a) Either A and B or B and C are sufficient to answer the
question
(b) Either A and B or A and C are sufficient to answer the
question
(c) Either A and C or B and C are sufficient to answer the
question
(d) A, B and C together are sufficient to answer the question
(e) Either only B or A and C together are sufficient to answer
the question
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Q99. Is |a × b| ≥ 10?
(A) a + b = –3
(B) a × b < 0
(C) a + 2.5b = 0 and both a & b are integers
(a) Only statement ‘A’ alone is sufficient to answer the
question
(b) Only statement ‘B’ alone is sufficient to answer the
question
(c) Only statement ‘C’ alone is sufficient to answer the
question
(d) Any of two statements are sufficient to answer the
question
(e) All three together are sufficient to answer the question
Q100. Three friends Sonu, Monu and Jonu live together. Age
of two of them is same. What is the age of Monu?
(A) Average age of all of them together is 32 years, 4 years
lesser than age of Sonu.
1
(B) Age of Monu is 333% lesser than age of Jonu and age of
Monu is 8 years lesser than the averge age of all three.
(C) Average age of 2 of them is 2 years lesser than average age
of all three together. Monu is youngest among them.
(a) Either A or B alone is sufficient to answer the question
(b) Either B or C alone is sufficient to answer the question
(c) Either A or C alone is sufficient to answer the question
(d) Any of the two statements are sufficient to answer the
question
(e) Either only A or B and C together is sufficient to answer
the question
Directions (101-105): Given below the table shows types of
interest offered by five banks, principal amount, time of
period and rate of interest. Some of the data is missing.
Calculate that according to information given in questions.
Bank
ICICI
SBI
YES
UCO
IDBI
Type of
interest
Compound
Simple
Compound
Compound
Simple
Principle
(Rs)
––
––
20000
––
10000
Time
(year)
––
4
3
2
––
Amount
(Rs)
––
26250
––
29160
–
Rate
15%
––
10%
––
6%
Q101. If the ratio of interest rate of IDBI to that of UCO is 3: 4,
then find the difference between principle invested in UCO
bank and amount obtained from IDBI, if time period for both
banks is same?
(a) 13800 Rs
(b) 12800 Rs
(c) 11800 Rs
(d) 13600 Rs
(e) 13900 Rs.
Q102. If rate of interest offered by SBI and Yes bank is same.
Then find principle invested in SBI is approximately what
percent of amount obtained from YES bank?
(a) 52%
(b) 59%
(c) 70%
(d) 65%
(e) 78%
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Q103. What is amount of interest obtained from ICICI bank, if
ratio of principle invested in ICICI bank to principal invested
in Yes bank is 7 : 5 and time period is one year less for ICICI
bank than time period of YES bank?
(a) 9020 Rs
(b) 9030 Rs
(c) 8030 Rs
(d) 7030 R
(e) 9080 Rs.
Q104. Principle invested in ICICI is 3000 more than principle
invested in UCO bank and both invested for same period of
time and UCO bank offered 8% rate of interest annually. If
amount obtained from ICICI is Rs. 32870 more than interest
obtained from UCO bank then find the principle invested in
UCO bank and ICICI bank?
(a) Rs 25000 & Rs 27000
(b) Rs 18000 & Rs 16000
(c) Rs 22000 & Rs 20000
(d) Rs 25000 & Rs 28000
(e) Rs 24000 & Rs. 28000
Q105. If ratio between rate of interest offered by SBI bank to
IDBI bank is 5 : 3 and ratio between time period is 2 : 1
respectively, then find the sum of principle invested in SBI
bank and amount obtained from IDBI bank?
(a) 27850 Rs
(b) 28850 Rs
(c) 29950 Rs
(d) 27950 Rs
(e) 31950 Rs.
Directions (106–110): The bar graph shows % of sweets
sold by a famous shop out of the total sweets that he prepared.
Answer the questions based on this information.
Note: All the units of sweets are given in KG’s unless
mentioned.
Barfi
Laddu's
100
90
80
70
60
50
40
30
20
10
0
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|
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Sunday
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Q106. Shopkeeper prepared 66 ⅔% more ‘Laddu’ on Sunday
than that of Saturday. Total number of Barfi prepared on
Sunday and Saturday are equal to total number of Laddu
prepared in these two days. Barfi prepared on both of days are
equal in quantity then find number of Laddu that remained
unsold on Sunday, given that difference between total sold
Barfi to total sold Laddu in these two days is 100kg.
(a) 2400
(b) 1200
(c) 900
(d) 1600
(e) None of these
Q107. Ratio of Laddu prepared on Friday and Laddu sold on
Monday together to total number of Laddu prepared on
Friday and Monday is 113 : 120. He earns profit of Rs. 20/kg
on selling Laddu and no loss on unsold Laddu. If total profit
earned on Monday is Rs. 11040 more than that of Friday on
selling Laddu then find quantity of Laddu prepared by him on
Monday.
(a) 1680 kg
(b) 1600 kg
(c) 1800 kg
(d) 1512 kg
(e) 1200 kg
Q108. Quantity of Laddu sold on Friday is equal to quantity of
Barfi sold on Monday. Calculate the quantity of Barfis
prepared on Friday, if he prepared 80 kg more Laddu than
Barfi on each day(Friday and Monday) and Barfi’s prepared
on Friday is 30% less than Barfi’s prepared on Monday.
(a) 1000 kg
(b) 1600 kg
(c) 1120 kg
(d) 1200 kg
(e) 1680 kg
Q109. If he earns a profit of Rs. 10/kg on selling each sweets
and loss of Rs. 10/800 gm on unsold items. Find his
approximate profit % on Saturday, if it cost Rs. 200/kg to
prepare each sweets and ratio of Laddu prepared to Barfi
prepared is 5 : 4 on Saturday.
(a) 2%
(b) 5%
(c) 10%
(d) 12%
(e) can’t be determined
Directions (111-112): The daily work of two men is equal to
that of 3 women or that of 4 youngsters. By employing 14 men,
12 women, and 12 youngsters a certain work can be finished
in 24 days.
Q111. If it is required to finish it in 14 days and as an
additional labor, only men are available, how many of them
will be required?
(a) 20
(b) 30
(c) 25
(d) 15
(e) None of these
1
Q112. If it is required to finish it in 195 days and as an
additional labor, only women and youngsters are available in
pairs, how many pairs of women and youngsters will be
required?
(a) 7
(b) 5
(c) 6
(d) 8
(e) None of these
Directions (113-115): A article is mark up above cost price
such that markup percent is double of the profit percent. If
1
discount is 12.5%, then profit percent increased by 33 3 %.
Q113. On selling 20 such article, profit is Rs.300. Find the M.P.
of each article.
(a) Rs.60
(b) Rs.160
(c) Rs.80
(d) Rs.240
(e) Rs.72
Q114. If shopkeeper cheat his customer by giving 20% less
quantity and reducing value of discount percentage by 20%
then find the new profit percent.
(a) 60%
(b) 75%
(c) 62.5%
(d) 80%
(e) 70%
Q110. Find the approximate average % of sweets sold on
Saturday, Sunday and Monday together if shopkeeper
prepared same quantity of Barfi on these days and ratio of
Barfi to laddu prepared in these 3 days is 4 : 3, 4 : 5 and 20 :
21 respectively.
(a) 85
(b) 76
(c) 60
(d) 68
(e) 65
16
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Q115. Initially shopkeeper have 20 articles. Out of 20, 7
articles damaged and remains unsold. Marked Price should he
labeled by how much percent more than cost price so that his
overall profit does not change neither his discount
percentage.
(a) 156%
(b) 146%
(c) 136%
(d) 120%
(e) 125%
Direction (116–120): Data given below shows farmers
producing three types of grains. Study the data carefully and
answer the following questions.
Farmers producing CORN is 20% more than farmers
producing RICE while number of farmers producing only
CORN and only RICE is same. Ratio between farmers
producing only WHEAT to farmers producing rice is 3 : 5.
Farmers producing WHEAT and CORN but not RICE is twice
of farmers producing only WHEAT and RICE. Number of
farmers producing only RICE is half of number of farmers
producing RICE.
Number of farmers producing RICE and CORN is 12% of total
number of farmers while number of farmers producing CORN
is 48% of total number of farmers. Number of farmers
producing all three grains is 16, which is 8 less than number
of farmers producing only RICE and WHEAT but not CORN.
Q116. Find number of farmers producing at least two types of
grain are what percent of number of farmers producing at
most one type of grain?
(a) 37.5%
(b) 43.75%
(c) 50%
(d) 56.25%
(e) 62.5%
Q117. Number of farmers producing Sugarcane is 25% more
than number of farmers producing WHEAT and CORN. Find
number of farmers producing Sugarcane?
(a) 80
(b) 50
(c) 60
(d) 45
(e) None of the given options
Q118. Find what percent of farmers producing WHEAT out of
total farmers?
(a) 24%
1
(b) 56 %
3
1
(c) 53 %
Q120. Find the ratio between number of farmers producing
only two types of grains to farmers producing all three grains?
(a) 27 : 4
(b) None of the given options
(c) 23 : 16
(d) 23 : 8
(e) 23 : 4
Directions (121-125): Pie-chart shown below shows
percentages of markers sold by six sellers.
Table shows ratio three type of marker out of total markers
sold by different sellers. Study the data carefully and solve the
following questions.
Total Markers sold = 15,000
A
12%
F
20%
B
15%
E
21%
C
18%
D
14%
Type of markers
A
B
C
D
E
F
X
4
3
9
6
3
4
Y
3
4
7
4
2
5
Z
2
3
9
5
1
3
Q121. Seller ‘A’ fixed his selling price of markers at 40%
above the cost price but at the time of selling he gave 40%,
20% and 10% discount on X, Y and Z respectively. Find the
total profit or loss percentage if cost price of all the markers is
same?
1
(a) 2 3 %
2
(b) 1 3 %
1
(c) 3 %
3
2
3
(d) 2 3 %
(d) 40%
(e) 48%
17
Q119. Find total number of farmers producing either only
RICE or only WHEAT?
(a) 120
(b) 132
(c) 280
(d) None of the given options
(e) 138
1
(e) 1 3 %
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Q122. Seller ‘E’ and ‘F’ keep the S.P. of each X, Y and Z markers
same and total S.P. of X, Y, Z sold by E is Rs.47250. Find the
total S.P. of all the markers sold by F if E kept the SP of each X,
Y, Z marker in the ratio 1 : 1.5 : 3.
(a) Rs.48250
(b) Rs.51250
(c) Rs.54520
(d) Rs.57520
(e) Rs.45500
Q126. A, C and D start working together but due to bad health
1
1
of A and D their efficiency decreased by 12 2 % and 33 3 %
respectively. Then find in how many hours total task will be
completed by these three?
1
(a) 22 hours
2
1
(b) 10 4 hours
1
(c) 12 4 hours
1
Q123. Seller ‘C’ sold all the markers for a certain sum and
1
there was a loss of 11 9 %. Had it been sold for Rs.9000 more,
(d) 9 hours
4
1
(e) 13 hours
4
1
there would have been a gain of 11 9 %. If seller ‘C’ wants to
earn 20% profit, then what would be the total S.P. of Y marker
if S.P. of each marker is in the ratio 2 : 3 : 4 respectively.
(a) Rs.13680
(b) Rs.12680
(c) Rs.13608
(d) Rs.12608
(e) None of these
Q124. There are two customers, Satish and Veer. Seller ‘B’
sells 60% of X marker to Satish, and remaining to Veer, B also
sells 40% of Y marker to Satish and remaining to veer. Find
the S.P. of each Y marker if Satish and Veer pays Rs.8370 and
Rs.9180 for X and Y marker respectively.
(a) Rs.10
(b) Rs.12
(c) Rs.14
(d) Rs.16
(e) Rs.18
Q125. Out of six sellers, which seller sells maximum number
of X type of marker?
(a) B
(b) C
(c) D
(d) F
(e) None of these
Direction (126 - 130): Given below bar graph show number
of hours taken by six persons to complete a task individually.
Read the data carefully and answer the questions:
100
90
Q127. E and F start working together on another task, while
F work with 25% less efficiency. E and F work for y hours and
remaining work complete by B in (y + 1) hours, if ratio of work
done by E and F together and by B alone is 2 : 1, then in how
many hours A will complete same task alone?
1
(a) 15 hours
2
1
(b) 13 2 hours
1
(c) 17 2 hours
1
(d) 11 2 hours
1
(e) 92 hours
Q128. If G can do 50 % more work in one hour as A can do in
one hour, while H can do 25% less work in one hour as B can
do in one hour. C start working alone and after some time he
left the task, if remaining task complete by G & H together in
23.5 hours more than C work alone. Then find total time in
which work completed?
(a) 32.5 hours
(b) 30.5 hours
(c) 28.5 hours
(d) 22.5 hours
(e) 16.5 hours
Q129. A, B, E and F work together in first hour, while C & D
together destroy the task (with same efficiency of completing
the task) in second hour. If this rotation continue till the total
work is completed. Find how many hours required to
complete the task?
17
(a) 55 hours
(b) 45
80
25
17
hours
25
177
70
(c) 40 245 hours
60
(d) 50
50
(e) 59 802 hours
40
30
20
10
0
A
18
B
C
D
E
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F
|
705
802
705
hours
Q130. If E work for 12 hours, B work for 35 hours, then find
in how many hours remaining work will be completed by C?
(a) 8 hours
(b) 10 hours
(c) 12 hours
(d) 15 hours
(e) 20 hours
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Directions (131-135): Table given below shows length of six
train, speed of train (meters/minutes), time taken by different
trains to cross different platform and length of each platform
is also given. Some of the data in table is missing, calculate the
missing data and answer the questions according to condition
given in questions.
Train
A
B
C
D
E
F
Length
of train
(m)
–––
180
–––
120
300
–––
Speed
(meters/
minutes)
750
–––
2000/3
–––
–––
1000
Time taken by
train to cross
platform (sec)
24
21.6
–––
–––
30
–––
Length of
platform
(m)
–––
–––
–––
240
–––
–––
Q135. Two train B and D moving in same direction, if speed
of smaller train is 54 km/hr and faster train crossed a man,
who sits in smaller train in 24 sec. Then find the speed of
faster train in meters/sec?
(a) 35/2 sec
(b) 45/2 sec
(c) 33/2 sec
(d) 43/2 sec
(e) 41/2 sec
Directions (136-140): Table given below shows number of
lectures taken on different days in four different weeks of a
month. Study the data carefully and answer the following
questions.
Q131. Train A running at its average speed crossed the
9
platform and it takes 911 sec to pass a man who is walking at
10 km/hr in the opposite direction to that of train A. Find the
length of platform?
(a) 300 m
(b) 150 m
(c) 180 m
(d) 200 m
(e) 225 m
Q132. Ratio of length of train C to train F is 3 : 5. If running in
opposite direction, both train crossed each other in 14.4 sec.
then find time taken by train C in crossing a platform which is
50m more than length of train of C. Also find length of train F?
(a) 53/2 sec, 250m
(b) 60/7 sec,200m
(c) 63/2 sec,250m
(d) 43/2 sec,180m
(e) 49/2 sec, 225 m
Q133. If train B and E crossed their respective platforms and
platform lengths are same as their respective train. Then find
in what time faster train crossed slower train if they are
running in same direction?
(a) 144 sec
(b) 134 sec
(c) 140 sec
(d) 240 sec
(e) 225 sec
Q134. Ratio of magnitude of time taken by train D in crossing
the respective platform to speed of train D (m/s) is 5 : 8. Then
find ratio of time taken by train D to train F in crossing a
platform whose length is 600 m? (Ratio of length of train D to
that of train F is 3:5)
(a) 5 : 8
(b) 5 : 6
(c) 7 : 5
(d) 2 : 5
(e) 5 : 3
19
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Lectures on
Monday and
Tuesday
each
Lectures on
Wednesday
and
Thursday
each
Lecture on
Friday and
Saturday
each
First
week
Second
week
Third
week
Fourth
week
Amount
paid for
one
lecture
per hour
8
—
6
9
Rs. 4,000
7
5
—
8
Rs. 6,000
6
6
8
10
Rs. 9,000
(i) There are no lectures on Sunday.
(ii) Month = First week + Second week + Third week +
Fourth week
Q136. A professor gets Rs. 2 lakhs to take all lectures on
Monday and Tuesday together in that month. If lectures are of
45-minute, 30-minute, 1 hour 40 minute and 40 minutes in
first, second, third and fourth week respectively then find
number of lectures taken on Monday of second week?
(a) 4
(b) 6
(c) 3
(d) 12
(e) 8
Q137. Find amount obtained by a professor who takes
lectures in whole fourth week having 40 minutes class on each
day of that month.
(a) 3.48 lakh
(b) 1.74 lakh
(c) 2.32 lakh
(d) 1.16 lakh
(e) None of these
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Q138. A professor takes lectures only on Wednesday. Find
lectures taken by him on third week is what percent more or
less than lectures taken by him on second week if he earns Rs.
1,21,500 by taking 45 minutes lecture on each week of that
month.
(a) 100%
(b) 60%
(c) 80%
(d) 20%
(e) 40%
Q139. Find the ratio between amount given to professor who
takes 45 minutes lecture on Wednesday of second week to
professor who takes 40 minutes lecture on Saturday of fourth
week.
(a) 1 : 2
(b) 3 : 8
(c) 1 : 3
(d) 3 : 5
(e) 8 : 27
Q140. Amount given to a professor who takes lecture on
whole second week [1 hour each] is 12.5% more than amount
given to that professor who takes lecture on whole first week
[45 minutes each]. Find numbers of lectures taken on Monday
of second week?
(a) 8
(b) 4
(c) 6
(d) 12
(e) None of these
Directions (141–145): Line chart given below shows
markup % above C.P. and discount % given on article X on five
different days while table given below shows ratio between
mark price of article X to mark price of article Y on these days.
Study the date carefully and answer the following questions.
Mark up %
Discount %
90%
80%
70%
60%
50%
40%
30%
20%
10%
0%
Days
Monday
Tuesday
Wednesday
Thursday
Friday
20
Q141. Ratio between cost price of article X on Monday to
Thursday is 4 : 5. Retailor earns Rs 75 profit on selling article
X on Thursday. Find profit earned on selling article Y on
Monday, if he give 35% discount on marked price of Y while
Cost price of article X and Y is same on Monday.
(a) Rs 465
(b) Rs 444
(c) Rs 525
(d) Rs 555
(e) Rs. 585
Q142. On selling article X and Y on Friday, retailor earns total
Rs 1430 profit. If selling price of article Y is Rs. 520 more than
selling price of article X and cost price of article X and Y is
same, then find discount percent offered on article Y.
1
(a) 33 3 %
2
(b) 16 %
3
2
(c) 66 %
3
(d) 25%
(e) 12.5%
Q143. Ratio of cost price of article X to article Y on Wednesday
is 3 : 4. Retailor earn 26% profit on selling article Y by giving
discount of Rs. 1008. Find the difference between selling price
of both articles.
(a) Rs. 900
(b) Rs. 540
(c) Rs. 360
(d) Rs. 1080
(e) Rs. 1260
Q144. Find profit percentage earned by retailor on selling
article Y on Tuesday at 40% discount if cost price of article Y
is 20% less than that of cost price of article X?
(a) 60%
(b) 25%
(c) 50%
(d) 75%
(e) 100%
Mark price of X : Mark price of Y
4:5
3:4
11 : 14
7:9
5:7
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Q145. If profit earned on selling article X on Monday is 176
then find the total profit earned by retailor on selling article X
on all five days. It is given that ratio between cost price of
article X on respective days starting from Monday to Friday is
1 : 2 : 3 : 4 : 5.
(a) None of the given as option
(b) Rs. 1432
(c) Rs. 1260
(d) Rs. 1302
(e) Rs. 1424
Directions (146-150): The following information is about
the production of bikes by 3 different companies from
Monday to Friday in a specific week. Read the information
carefully and answer the following question:
The total production by 3 companies on Monday was 540 out
1
of which 33 3 % bikes were produced by Hero. The number of
bikes produced by Bajaj on Monday are less than the bikes
produced by Hero on Monday by the same extent as the
number of bikes produced by Honda on Monday is more than
the bikes produced by Hero on Monday. The difference
between bikes produced by Bajaj and Honda on Monday is 40.
150 bikes are produced by Hero on Tuesday, which is 100 less
than the bikes produced by the same company on Wednesday.
A total of 910 bikes were produced by Hero from Monday to
Friday. The ratio between bikes produced by Hero on
Thursday to bikes produced by the same company on Friday
is 5 : 6.
220 bikes were produced by Bajaj on Tuesday, which is 80
less than the bikes produced by Honda on Wednesday. A total
of 570 bikes were produced on Tuesday, which is 76% of the
total bikes produced on Wednesday. The number of bikes
2
produced by Honda on Thursday is 66 3 % more than bikes
produced by Hero on the same day. Total 580 bikes were
produced on Thursday. The number of bikes produced by
Honda on Friday is same as that on Monday. 140 bikes were
produced by Bajaj on Friday.
Q146. Find the ratio between total bikes produced on Monday
to that on Wednesday.
(a) 18 : 29
(b) 18 : 25
(c) 18 : 31
(d) 3 : 5
(e) None of these
Q147. Find the total number of bikes produced by Bajaj from
Monday to Friday.
(a) 900
(b) 980
(c) 950
(d) 960
(e) None of these
21
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Q148. Find the average number of bikes produced per day by
Honda from Monday to Friday. (approximate)
(a) 250
(b) 220
(c) 270
(d) 240
(e) 230
Q149. On which pair of days out of the following, the number
of bikes produced by Hero is the same ?
(a) Tuesday and Wednesday
(b) Wednesday and Thursday
(c) Tuesday and Thursday
(d) Monday and Wednesday
(e) Monday and Tuesday
Q150. On which day the total number of bikes produced was
the maximum ?
(a) Monday
(b) Tuesday
(c) Wednesday
(d) Thursday
(e) Friday
Q151. Ram spends Rs. 8000 for 4 years in scheme ‘A’ which
offers R% p.a. at S.I. and Rs. 10,000 for 2 years in scheme ‘B’
which offers R% p.a. at C.I. He got same amount from both
schemes. Find value of ‘R’.
I. 20%
II. 50%
III. 100%
(a) Either I or II
(b) Either I or III
(c) Either II or III
(d) Any one of I, II and III
(e) Only I
Q152. A retailer bought some eggs at Rs. 10 each. Out of these,
some are rotten, and some are fresh. He sold fresh eggs at 50%
profit and rotten eggs at 20% loss. In total he earns 8% profit.
Find the probability of selecting one fresh egg from total eggs.
1
(a) 9
1
(b) 2
(c)
(d)
4
5
3
5
2
(e) 5
Q153. Two equilateral triangles and two parallelograms are
cut down from a hexagon. If area of one equilateral triangle is
4√3 cm² then find the height of parallelogram?
(a) 2√3 cm
(b) √3 cm
(c) 4√3 cm
(d) 3√3 cm
(e) 8√3 cm
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Q154. Work done by A, B, C and D is in arithmetic progression.
C worked for 9 days and completed 30% work. A and B
worked for 2 days and 3 days only. Find for how many days
‘D’ work to complete the remaining work if all four together
can complete that work in 4 days.
(a) 12 days
(b) 10 days
(c) 8 days
(d) 6 days
(e) 4 days
(a) 12100
(b) 14400
(c) 13800
(d) 15000
(e) None of these
Directions (155-156): Pipe A is 33 ⅓% more efficient than
pipe B. Pipe C can empty the tank at the rate of 40 litres/hour.
18 11 % less of their respective efficiency respectively and B
Q155. Mohit opened pipe A and pipe B for
13
7
hours and the
tank was filled upto10 liters less than 75% of its capacity.
After that he also opened the pipe C and tank was filled in
1 ℎ𝑜𝑢𝑟. Find the time in which pipe C alone could empty the
tank.
(a) 8 hours
(b) 9 hours
(c) 9.5 hours
(d) 8.5 hours
(e) Can’t be determined
Q156. When these pipes were connected to a different tank.
Initially, pipe A and B were opened, till the tank was filled 10
liters more water than 75% capacity of tank. After that all the
pipes were opened, and the remaining tank was filled in 1
hour. Find the capacity of tank. Given that pipe A will take 4
hours to fill the tank.
(a) 600 litres
(b) 160 litres
(c) 360 litres
(d) 180 litres
(e) can’t be determined
Q157. Speed of current is 10 km/hr and speed of a motor boat
is 80% more than speed of current. Motor boat travels 280 km
downstream with its usual speed, after that it’s increased
speed by ‘s’ kmph and travelled for another 280 km then it
returns and covers 560 km in upstream. If boat complete
whole journey downstream to upstream in 45 hr, then find the
value of ‘s’?
(a) 10 km/hr
(b) 8 km/hr
(c) 6 km/hr
(d) 12 km/hr
(e) 4 km/hr
Q158. A, B and C enter into a partnership and invested some
amount. After one year A double its investment, B increase its
1
investment by 33 3 % and C increase its investment by 20%.
In the third year A and B withdraw their investments and D
joins the partnership with C. After three year they got profit
in the ratio of 12 : 14 : 17 : 8 (A : B : C : D). If difference between
initial investment of B and C is 1150. Then Find out the total
initial investment made by A and D together?
22
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Q159. A and B together can finish first work in
1
11 4 𝑑𝑎𝑦𝑠 while A alone can do it in 25 days. Both work on
2
second work for ‘x’ days and (x + 4) days with 22 9 % more and
2
gets Rs 180 more wages than that of A out of total wages Rs
4140. Find the time taken by both together to complete the
second work if they work with their original efficiency?
4
(a) 13 5 𝑑𝑎𝑦𝑠
1
(b) 12 𝑑𝑎𝑦𝑠
2
2
(c) 16 𝑑𝑎𝑦𝑠
3
1
(d) 13 𝑑𝑎𝑦𝑠
3
(e) 14 𝑑𝑎𝑦𝑠
Q160. A shopkeeper raised the marked price of an article by
125% and allows three successive discounts of 25%, 20% and
1
119 % on new marked price and make a profit of 25%. If
1
shopkeeper allows three successive discounts of 33 3 %,
2
1
16 3 % and 122 % on new marked price, he would have Rs 255
less profit than earlier, then find cost price of that article?
(a) 2184 Rs.
(b) 2244 Rs.
(c) 2284 Rs.
(d) 2304 Rs.
(e) 2364 Rs.
Directions (161-165): The following table shows different
plans offered by a lender, type of interest and rates of interest
applicable during first, second and third years.
(Note: Some values are missing, you need to calculate those
values if required.)
Rate of Interest
Type of
Plans
First
Second
Third
Interest
Year
Year
Year
A
Simple
2
2
6 3%
3 3%
Interest
B
Compound
1
6 %
4
Interest
C
Simple
3
1
8 %
5 %
4
4
Interest
D
Compound
1
3
7 2%
4 4%
Interest
4
3
E
Simple
5 %
4 %
5
5
Interest
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Q161. If two persons borrows an equal amount of Rs.12000
under plan B and plan E respectively and rate of interest for
the first year under plan B and D is same, then what is the
difference between second year’s interests alone paid by each
of them?
(a) Rs.105.25
(b) Rs.110.25
(c) Rs.115.25
(d) Rs.120.25
(e) Cannot be determined
Directions (166-170) : A new party XYZ, participated in
election in 5 constituencies (A, B, C, D & E). Pie chart shows %
distribution of party’s total votes scored by its 5 candidates
and line graph shows, % of votes scored by winner candidates
out of total votes polled in these 5 constituencies.
Note: All the votes must be calculated in nearest hundreds.
TOTAL VOTES = 300000
A
10%
E
12.5%
Q162. A person borrows Rs.20480 under plan C. After
completion of the loan tenure of three years under plan C, he
extends the tenure for further two years under plan D on the
amount payable at that time. He settles his loan by paying
Rs.27778. What is the rate of interest for the second year
under plan D if rate of interest for the third year under plan C
and D is same?
3
(a) 5 %
B
25%
D
15%
4
1
(b) 5 4%
C
37.5%
1
(c) 6 4%
3
(d) 4 4%
3
(e) 6 4%
70
Q163. If the amounts borrowed by a person under plan B and
C are in the ratio 16 : 13 and rate of interest applicable during
the first year under plan B and D is same, then what is ratio of
interests payable under these plans at the end of second year.
(a) 5 : 6
(b) 3 : 5
(c) 3 : 4
(d) 5 : 4
(e) None of these
Q164. The lender decides to offer a fixed rate of interest at
2
6 3% per year under plan C. By how much percent the interest
payable will increase from the interest payable previously
under the old plan for the period of three years if rate of
interest for the third year under old plan C and plan D is same?
1
(a) 6 3%
1
(b) 6 4%
2
(c) 6 %
3
2
(d) 6 5%
(e) Cannot be determined
Q165. Rates of interest for the first year under plan A and E
2
3
are 8 3% and 7 5% respectively. A person borrows a total of
Rs.30000 partially under plan A and E and pays a total interest
of Rs.5540 at the end of third year. How much amount does he
borrow under plan A?
(a) Rs.14000
(b) Rs.18000
(c) Rs.16000
(d) Rs.12000
(e) Rs.20000
23
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60
50
40
30
20
10
0
A
B
C
D
E
Q166. Candidate of party XYZ is winner from constituency B
and defeated runner-up candidate by 7000 votes. None of the
candidates from this constituency obtained vote less than
12000, then calculate maximum possible number of
candidates from constituency B.
(a) 6
(b) 7
(c) 15
(d) 13
(e) can’t be determined
Q167. Party XYZ wins from constituency C. In constituency E
total votes polled are 45000 less than that of constituency C.
Find from how many votes the candidate of party XYZ lost in
constituency E?
(a) 79500
(b) 91500
(c) 26500
(d) 27000
(e) 0
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Q168. Out of constituency A and D, one of the seats was won
by party XYZ. In lost constituency their candidate was runner
up by 15000 votes. Find maximum possible difference
between total votes polled in both of these constituencies.
(a) 50000
(b) 37500
(c) 25000
(d) either (a) or (b)
(e) either (b) or (c)
Q169. In constituencies A & E, both candidates of party XYZ
are runner up, and lost by same number of votes. If the votes
polled in E is 180000, find approximate number of votes
polled in constituency A.
(a) 350000
(b) 365000
(c) 180000
(d) 100000
(e) None of these
Q170. Which of the following condition is never possible?
(i) constituency B and C have equal number of votes polled
given that party wins from C.
(ii) candidate of party XYZ lost by 28000 voters from
constituency E.
(iii) constituency A and B have equal number of votes polled.
(iv) winner candidate of constituencies B and D got equal
number of votes.
(a) only (i)
(b) Only (i) and (iii)
(c) (i), (ii) and (iv)
(d) (i) and (ii)
(e) All of these are possible
Directions (171-175): Data show the different kind of solids
in a toy shop. Shopkeeper or (toymaker) makes different
types of toys by joining these solids. Some values are missing,
you have to calculate these values if required to answer the
question.
Diameter Length Breadth Height
Cylinder
–
–
–
12
Cube
–
–
–
–
Cuboid
–
24
–
10
Cone
14
–
–
–
Sphere
21
–
–
–
Hemisphere
−
–
–
–
Q171. A toymaker makes a toy in which a cone is mounted on
the base of a hemisphere. If the total surface area of the toy is
858 cm² then find the volume of the toy?
2
(a) 1950 3 cm²
2
(b) 1250 3 cm²
(c) 1400 cm²
(d) 1500 cm²
(e) None of these
24
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Q172. Toymaker mounted the cube on the cylinder such that
cylinder top is exactly in the middle of the face of the cube.
Find the total surface of the toy formed, if the height of formed
11
toy is 4 𝑡ℎ of the height of cylinder and curved surface area of
cylinder is 66 times the height of cylinder.
(a) 3125 cm²
(b) 2794.5 cm²
(c) 4112 cm²
(d) 5123 cm²
(e) 3438 𝑐𝑚2
Q173. If given sphere is cut into two hemisphere and these
hemispheres are mounted on both ends of the cylinder, then
find out the ratio of volumes of toy formed by joining both
hemispheres on cylinder, cylinder and sphere.
(a) 7 : 6 : 13
(b) 6 : 13 : 7
(c) 13 : 6 : 7
(d) 13 : 7 : 6
(e) None of these
Q174. Volume of the cuboid is approximately what percent
more or less than the volume of cone if slant height of cone is
25 cm and the breadth of the cuboid is 25% of the height of
cone.
(a) 7%
(b) 11%
(c) 14%
(d) 17%
(e) 21%
Q175. A solid right circular cylinder has radius r and height
5r. A solid right circular cone is carved out from one end of the
base of cylinder. If base radius of cone is r and height is 2√2 r
then, find the ratio between total surface area of cone to the
total surface area of remaining part of cylinder.
(a) 3 : 5
(b) 4 : 7
(c) 2 : 7
(d) 3 : 4
(e) 1 : 3
Directions (176-180): Table given below shows the amount
paid, mileage and time (in hours) for which five taxis are hired
by a business man at the rate of Rs. 5/km plus the cost of
petrol at Rs. 60/Litre.
Taxi
P
Q
R
S
T
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Amount paid
(in Rs.)
2640
3500
2592
4500
2925
|
Mileage (in
km/Litre)
10
12
15
18
14
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Time
(hours)
16
25
18
30
15
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Q176. Amount paid to taxi Q and S together will how much
increase, if mileages of both taxis are decreased by 2km/Litre
(distance covered will be same.
(a) Rs.470
(b) Rs.475
(c) Rs.575
(d) Rs.674
(e) Rs.325
Q177. Average speed of taxi Q is what percent less/more than
the average speed of taxi P?
1
(a) 13 %
3
2
(b) 12 %
3
(c)10%
2
(d) 6 3 %
2
(e) 16 3 %
Q178. If taxi S travel his journey in two equal halves with two
different speed having ratio 3:2 then find the time for which it
travels with higher speed?
(a) 12 hours
(b) 10 hours
(c) 16 hours
(d) 18 hours
(e) 14 hours
Q179. What is the ratio of the average speed of the taxi R to
that of taxi T?
(a) 21:16
(b) 12:23
(c) 14:17
(d) 16:21
(e) None of these
Q180. Find the average of the total distance covered by all five
taxis?
(a) 314.6 km
(b) 298.4 km
(c) 412 km
(d) 344.6 km
(e) 346.6 km
Q181. A mixture contains milk and water in the ratio of 9 : 2.
44 lit of mixture is taken out and 12 lit of water is added to it,
such that ratio of milk to water becomes 3:1. Now another
mixture of 64 lit having milk and water in ratio of 3:5 is added
to it. Find ratio of milk to water in the final mixture?
(a) 32 : 17
(b) 34 : 19
(c) 7 : 4
(d) 24 : 11
(e) 33 : 19
25
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Q182. Ratio of present age of Simmi & Rimmi is 4 : 3 while
ratio of age of Simmi 6 years later to present age of Rina is 3 :
1. Ratio of present age of Rimmi to present age of Rina is 2 : 1.
Find average of present age of all the three (in years).
(a) 37
(b) 36
(c) 32
(d) 34
(e) 38
Q183. A shopkeeper sells an article at Rs. 720 by making a
profit of x% and on interchanging selling price with cost price,
it would have a loss of y%. Find the selling price if an article
was sold at y% profit whose cost price is Rs.720 (given x : y =
9 : 7)
(a) Rs. 810
(b) Rs. 800
(c) Rs. 1100
(d) Rs. 880
(e) Rs. 660
Q184. There are two rectangular fields of same area. The
length of first rectangular field is a% less than the length of
the second field and breadth of the first field is (4a)% greater
than the breadth of the second field. Find the value of ‘a’ if a is
non-zero number.
(a) 60
(b) 75
(c) 80
(d) 90
(e) None of these
Q185. A sum is divided between Aman and Vikash in the ratio
of 3 : 5. Aman purchased a motorbike from his money which
depreciates 20% per annum while Vikash invested his
amount in a scheme which offers compound interest at 20%
per annum. By what percentage, the total sum will change
after two years?
(a) 12%
(b) 14%
(c) 16%
(d) 18%
(e) 20%
Directions (186-190): Read the following table and line
graph carefully and answer the following questions.
Following table shows the time taken by five persons to
complete a work on Monday and Ratio of Time taken by these
five persons to complete the work on Monday to the time
taken to complete the work on Wednesday is also given.
Line graph shows the efficiency (as a percentage) of these five
persons on Tuesday with respect to that on Monday.
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Person
Time taken
to
complete
the work
on Monday
Ratio of Time taken to
complete the work on
Monday to the time taken to
complete the work on
Wednesday
Gaurav
25 min.
5:4
Abhishek
20 min.
4:5
Shailesh
50 min.
10 : 7
Neeraj
10 min.
5 : 13
Arunoday
150 min.
3:5
Q188. On Tuesday, Abhishek, Shailesh and Neeraj work in a
rotation in this order to complete the job with only 1 person
working in a minute. They earned a total of 875 Rs. Find the
share of Shailesh.
(a) 41 Rs.
(b) 31 Rs.
(c) 51 Rs.
(d) 49 Rs.
(e) None of these
Q189. On Tuesday, Aman who is half as efficient as Shailesh,
worked for 50 minutes on the same day then he left. In how
many minutes Neeraj and Abhishek together will complete
the remaining work ?
120
2
(a) 5 9 mins.
80
(b) 4 7 mins.
percentage →
100
3
3
(c) 5 mins.
60
7
1
(d) 4 7 mins.
40
5
(e) 5 7 mins.
20
0
Person →
Q186. Gaurav, Abhishek and Neeraj work in a rotation to
complete the job on Tuesday with only 1 person working in a
minute. Who should start the job so that the job is completed
in the least possible time?
(a) Gaurav
(b) Abhishek
(c) Neeraj
(d) Any one of three
(e) Can’t determine
Q187. On Tuesday, Gaurav and Arunoday started the work
and they worked for 5 minutes then Gaurav is replaced by
Abhishek. In how many minutes Abhishek and Arunoday
complete the remaining work ?
3
(a) 20 7 min.
(b) 21
(c) 21
4
21
5
21
4
min.
min.
(d) 20 17 min.
(e) None of these
26
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Q190. On Wednesday, all of them started the work together.
After working for 2 minutes Gaurav left. All except Gaurav
worked for another 3 minutes and then all left except
Arunoday. In how much time Arunoday will complete the
remaining work? (find the approximate value)
(a) 86 minutes
(b) 81 minutes
(c) 96 minutes
(d) 56 minutes
(e) 79 minutes
Directions (191-195): Each of the following question is
followed by two quantities I, and II. You have to determine the
value of the quantities using the information provided and
accordingly compare the quantities. Mark your answer as per
the instruction set provided below.
(a) Quantity I>Quantity II
(b) Quantity I≥Quantity II
(c) Quantity I<Quantity II
(d) Quantity I≤Quantity II
(e) Quantity I=Quantity II or no relation
Q191. Quantity I: Time taken by Prabhas to cover a distance
of 200 km (without stoppage) by car through which he covers
a distance of 100 km in 1 hour when he stops to get fuel for 10
minutes.
Quantity II: Time taken by Nagarjuna to travel 20 km
downstream if speed of boat and speed of stream be 7 kmph
& 3 kmph respectively.
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Q192. There are 3 red and 5 blue balls in the urn.
Quantity I: the probability of drawing 1 red ball and 2 blue
balls.
Quantity II: the probability of drawing 2 red balls and 1 blue
ball.
Q198. A bag contains total 12 balls in which there are 5 green
balls and rest are blue and red balls. What is difference
between blue & red balls.
Statement I . If one ball taken out from bag probability of
7
being either red or blue is 12.
Statement II. If two balls taken out from bag probability of
Q193. In how many ways can 4 boys or 5 girls be selected?
Quantity I: there are 20 persons (boys and girls only) in the
group out of which 12 are boys.
Quantity I: the group comprises 10 boys and 10 girls.
Q194. Quantity I: The length & breadth of a rectangle of
perimeter 48 cm are in ratio 5:3. Find Area?
Quantity II: 135
1
1
𝑎
𝑏
Q195. + =?
Quantity I: a & b are roots of equation 𝑥 2 + 𝑥 − 6 = 0
Quantity II: 𝑎2 + 𝑏 2 = 20; 𝑎𝑏 = 8
Direction (196-200): Following are the questions based on
two statements and answer the following based on the given
statements.
Q196. What will be respective ratio of saving of Veer &
Deepak.
Statement I. Income of Veer is 4% less than that of Sameer
and also expenditure of Veer is 12.5% less than that of
3
Sameer. Deepak spend 5 th of his income.
Statement II. Sameer save Rs. 7000 & Veer save Rs. 7400.
Income of Deepak is Rs. 1000 more than that of Sameer.
(a) Only statement I is sufficient
(b) Only statement II is sufficient
(c) Statement I and II both together is sufficient
(d) Either statement I or Statement II alone is sufficient
(e) Neither statement I nor statement II is sufficient
1
being either red or blue is 6.
(a) Only statement II is sufficient
(b) Either statement I or Statement II alone is sufficient
(c) Statement I and II both together is sufficient
(d) Only statement I is sufficient
(e) Neither statement I nor statement II is sufficient
Q199. Side of square is 3.5 cm more than radius of circle.
What will be area of square?
Statement I. Difference between circumference and diameter
of circle is 45 cm.
Statement II. Radius of circle is 50% more than breadth of
rectangle whose length is 15 cm. Ratio of circumference of
circle & perimeter of rectangle is 3 : 2.
(a) Only statement II is sufficient
(b) Either statement I or Statement II alone is sufficient
(c) Statement I and II both together is sufficient
(d) Only statement I is sufficient
(e) Neither statement I nor statement II is sufficient
Q200. What will be length of train A?
Statement I. Relative speed of train A & B is 10 meters/sec
when both running in same direction and length of train B is
240 (Speed of train B is more than speed of train A).
Statement II. Train B cross a pole in 8 sec and cross train A in
12 sec running in opposite direction.
(a) Only statement II is sufficient
(b) Either statement I or Statement II alone is sufficient
(c) Neither statement I nor statement II is sufficient
(d) Only statement I is sufficient
(e) Statement I and II both together is sufficient
Q197. What will be cost price of article, which marked 40%
above.
Statement I. If article sold 25% discount profit will be Rs. 50.
2
Statement II. If article sold two successive discounts of 14 %
7
and 10% profit will be Rs. 80.
(a) Only statement I is sufficient
(b) Either statement I or Statement II alone is sufficient
(c) Statement I and II both together is sufficient
(d) Only statement II is sufficient
(e) Neither statement I nor statement II is sufficient
27
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Directions (201-215): What will come in place of
question mark (?) in the following questions.
Q201. √5776 - √1444 + √729 =43 + ?
(a) 25
(b) 20
(c) 26
(d) 24
(e) 22
Q208. 73823–34156+4756+6758–9849 = 41499–160?
(a) 5
(b) 7
(c) 4
(d) 8
(e) 6
5599
3773
(a) 44
(b) 46
(c) 48
(d) 50
(e) 52
Q202. 78 ×26÷6 +1262= 1311 + (?)2
(a) 17
(b) 22
(c) 15
(d) 13
(e) 19
1
7
(a) 2
(b) 1
(c) 0
(d) 3
(e) 4
3840
1440
1330
Q211. √ 60 + 40 − 70 = ?
Q204. 42.5×15 +37.5× 25= 1420 + ?
(a) 145
(b) 165
(c) 155
(d) 170
(e) 185
(a) 10
(b) 9
(c) 8
(d) 7
(e) 11
Q205. 2450 +3760 -3830 =6000 - ?
(a) 3610
(b) 3620
(c) 3580
(d) 3600
(e) 3520
Q212. 25×18+
𝑜𝑓25
20
Q210. 84 × 4 ÷ 212 +? = 147 × 21 − 21
Q203.1484÷28 + 1462÷34 -12×7= ?
(a) 12
(b) 14
(c) 18
(d) 16
(e) 20
4
88
Q209. 1331 × 2036 × 49 =? −62
4200
40
−
525
105
= 740 − ?
(a) 200
(b) 220
(c) 190
(d) 170
(e) 150
3
1
(a) 50
(b) 45
(c) 35
(d) 30
(e) 40
Q213. 3845+4380+2640 − 5965 = (?)2
(a) 75
(b) 60
(c) 80
(d) 70
(e) 72
Q207. 55% 𝑜𝑓 900 + 70% 𝑜𝑓 1050 = ? % 𝑜𝑓 3000
(a) 41
(b) 42
(c) 43
(d) 44
(e) 45
Q214. 400 ÷ 20 × 35 + 6666 ÷ 33+ ? = 1100
(a) 180
(b) 198
(c) 195
(d) 205
(e) 200
Q206. (5 64 ) ÷ (432 − 202 + 7 𝑜𝑓 21) × (82) =? 𝑜𝑓 64
28
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Q215. 28× 14.5+1680÷15+445=1000 -?
(a) 27
(b) 37
(c) 47
(d) 50
(e) 40
Q222.
(a) 96
(b) 126
(c) 156
(d) 196
(e) 84
Directions (216-230): what approximate value will come
in place of question (?) mark:
Q216. 129.89% of 1199.82 + 1249.78 ÷ 49.98 × 30.012 = ?
(a) 2210
(b) 2380
(c) 2310
(d) 2530
(e) 2460
Q217. 155.9 ÷ √168.81+ (2.98)² × 39.89 = ? % of 599.92
(a) 62
(b) 78
(c) 84
(d) 52
(e) 68
Q218. √80.98 × 36.01 + 679.81 ÷ 17.01= ? + (511.98)1/3
(a) 86
(b) 78
(c) 94
(d) 52
(e) 66
Q219. 1599.85% of 139.89 + ? % of 1599.83 = 72.01 ×
39.81
(a) 20
(b) 32
(c) 60
(d) 50
(e) 40
Q220. (17.012)² + (21.89)² + (8.01)² + ? = 1749.821 –
820.01 + 2210.01
(a) 2208
(b) 2256
(c) 2601
(d) 2303
(e) 2373
Q221. 307.89 + 671.93 – 39.87% of ? + 79.89% of 354.93
= (27.87)²
(a) 1200
(b) 1175
(c) 1225
(d) 1250
(e) 1280
29
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177.8 + ?
|
7.98
+24.89×41.87 –15.98 % of 400 = (31.89)2
Q223. √1295.96 + √2024.93 + √1520.97 – √? = 12.93 %
of 899.98
(a) 5
(b) 7
(c) 13
(d) 16
(e) 9
Q224. 349.89 +
(a) 196
(b) 441
(c) 400
(d) 529
(e) 625
55.98 ×239.89
13.86
+ √? = (10.98)3
Q225. 31.96 × 34.89 +
39.98 % of 3304.98
(a) 800
(b) 700
(c) 900
(d) 1000
(e) 950
√960.89 + 18.98 % of ? =
Q226. 1782.011 ÷ 53.99 + 455.889 – 2346.011 × 1.011 = ?
× 2.93
(a) –629
(b) –619
(c) 629
(d) 619
(e) –609
Q227. (574.99 + 7511.11 – 2768.91) ÷ (76.1 × 0.98 +
674.976 – 342.001) = √?
(a) 529
(b) 49
(c) 169
(d) 289
(e) 729
1
Q228.
[(√3843.9 × 9.09) ÷ (26.99)3 ] × 23.012 =?2 +
336.97
(a) 33
(b) 23
(c) 27
(d) 37
(e) 43
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Q229. √(95.99) × 12.01 ÷ 17.9 + 25.899– 9.011 =(64.9–
?)% of 35.88
(a) 50
(b) 35
(c) 30
(d) 40
(e) 20
Q236. 1229.99 + 2120.09 − 3049.987 =?
(a) 300
(b) 100
(c) 200
(d) 500
(e) 400
Q237. √√(99.99 + 104.99 × 5 =?÷ 8.989
Q230. 11.9 × √224.89 + 1212.09 – (1053.11 ÷ 8.9) = ?
(a) 1,275
(b) 1,225
(c) 1,175
(d) 1,255
(e) 1,245
Directions (231-245): What approximate value will come
in place of question mark (?) in the following questions.
(You are not expected to find the exact value)
Q231. 42.022% 𝑜𝑓 350.09 − 28.04% 𝑜𝑓 399.999 =?
(a) 40
(b) 35
(c) 45
(d) 50
(e) 30
Q232. √(123.09 + 465.05) ÷ 11.99+? = 240.02 ÷ 1.989
(a) 93
(b) 143
(c) 133
(d) 113
(e) 123
Q233. (15.99)2 − 14.04 × 8.99+? = 154.999
(a) 30
(b) 45
(c) 35
(d) 20
(e) 25
(a) 55
(b) 15
(c) 25
(d) 35
(e) 45
Q238. 35.99 × 4.98 − 1199.99 ÷ 7.99 =?
(a) 20
(b) 50
(c) 40
(d) 30
(e) 10
Q239. ?2 + 60% 𝑜𝑓 239.99 = 55% 𝑜𝑓 320.02 + 3.98
(a) 8
(b) 6
(c) 4
(d) 16
(e) 14
Q240. 524.90 + 125.05 =? × 9.99
(a) 85
(b) 75
(c) 65
(d) 55
(e) 45
Q241. √144.04 × 15% 𝑜𝑓 120.09 =? −54.99 × 3.03
(a) 401
(b) 431
(c) 341
(d) 471
(e) 381
Q234. 62.02% 𝑜𝑓 249.99 − 19.99% 𝑜𝑓 105.05−? = 110
(a) 24
(b) 16
(c) 28
(d) 34
(e) 20
Q242. 13.03 × 7+? = 30.03% 𝑜𝑓 349.99
(a) 14
(b) 18
(c) 8
(d) 20
(e) 6
Q235. 44.98% 𝑜𝑓 220.09 + 30.03% 𝑜𝑓 160.06 =?2 + 2.99
(a) 32
(b) 28
(c) 12
(d) 22
(e) 18
Q243. 32.01% of 600.02 – 19.99% of 400.04+?=859.99÷2
(a) 358
(b) 258
(c) 288
(d) 318
(e) 338
30
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141
Q251. 810, 820, 832, 868, 1012, 1732, 6052
(a) 6052
(b) 810
(c) 868
(d) 832
(e) 1732
279.89
Q244. 20.09 + 39.99 − √? = 10.01
(a) 36
(b) 16
(c) 4
(d) 64
(e) 100
Q245. 8.98 × 60.02 − 19.992 + 10.01% 𝑜𝑓 130.09 =?
(a) 123
(b) 93
(c) 153
(d) 173
(e) 113
Directions (246-260): In each of these questions a
number series is given. In each series only one number, if
any, is wrong. Find out the wrong number.
Q246. 28, 14, 14, 22, 42, 105, 315
(a) 28
(b) 42
(c) 315
(d) 22
(e) 105
Q252. 1024, 350, 832, 508, 704, 604, 640
(a) 1024
(b) 640
(c) 704
(d) 350
(e) 508
Q253. 190, 210, 266, 358, 486, 646, 850
(a) 646
(b) 850
(c) 486
(d) 190
(e) 210
Q254. 15, 50, 160, 370, 709, 1208, 1904
(a) 15
(b) 50
(c) 370
(d) 1208
(e) 15
Q247. 5, 7, 13, 25, 47, 75, 117
(a) 5
(b) 7
(c) 75
(d) 117
(e) 47
Q248. 288000, 24000, 3600, 300, 50, 12.5, 6.25
(a) 24000
(b) 50
(c) 12.5
(d) 3600
(e) 6.25
Q255. 120, 170, 251, 367, 522, 720, 990
(a) 120
(b) 990
(c) 522
(d) 367
(e) 251
Q256. 55, 120, 210, 338, 517, 760, 1090
(a) 120
(b) 1090
(c) 760
(d) 55
(e) 338
Q249. 120, 125, 136, 149, 166, 185, 208
(a) 120
(b) 166
(c) 149
(d) 185
(e) 208
Q250. 205, 214, 186, 250, 125, 341, −2
(a) 205
(b) 214
(c) 250
(d) 125
(e) -2
31
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Q257. 110, 140, 240, 261, 365, 380, 492
(a) 240
(b) 380
(c) 492
(d) 140
(e) 110
Q267. I. 𝑥 2 − 2𝑥 − 15 = 0
II. 𝑦 2 − 15𝑦 + 56 = 0
Q268. I. 10𝑥 2 + 19𝑥 + 7 = 0
II. 5𝑦 2 + 16𝑦 + 12 = 0
Q269. I. 𝑥 2 − 20𝑥 + 75 = 0
II. 𝑦 2 + 19𝑦 + 84 = 0
Q258. 105, 106, 123, 154, 197, 255, 327
(a) 197
(b) 105
(c) 154
(d) 255
(e) 123
Q270. I. 𝑥 2 − 9𝑥 − 22 = 0
II. 𝑦 2 − 17𝑦 + 66 = 0
Q271. I. 4𝑥 2 + 19𝑥 + 15 = 0
II. 8𝑦 2 + 10𝑦 + 3 = 0
Q259. 1, 329, 638, 911, 1130, 1277, 1334
(a) 1
(b) 1334
(c) 911
(d) 1277
(e) 638
Q272. I. 𝑥 2 − 18𝑥 + 56 = 0
II. 𝑦 2 + 4𝑦 − 32 = 0
Q273. I. 𝑥 2 + 14𝑥 − 72 = 0
II. 𝑦 2 − 13 + 36 = 0
Q260. 2100, 2136, 1990, 2316, 1740, 2640, 1344
(a) 2100
(b) 1990
(c) 2316
(d) 1740
(e) 2640
Q274. I. 𝑥 2 − 92 = 122
II. 𝑦 3 = 3375
5
Q275. I.
𝑥2
28
3
=
𝑥2
7
II. 11𝑦 + (7 × 6) = 97
Direction (261-275): Given below in each question two
quadratic equations are given. Please solve each quantity
and compare both of them and answer accordingly from
the following options.
(a) x>y
(b) y> x
(c) x ≥y
(d) x ≤y
(e) x =y or No relation can’t be established.
Q261. I. 2𝑥 2 + 𝑥 − 6 = 0
II. 𝑦 2 + 6𝑦 + 9 = 0
Q276.
I. 𝑥 2 − 22𝑥 + 72 = 0
II. 𝑦 2 + 11𝑦 + 30 = 0
Q262. I. 𝑥 2 − 4𝑥 + 4 = 0
II. 𝑦 2 − 10𝑦 + 16 = 0
Q277.
I. 𝑥 2 − 23𝑥 + 120 = 0
II. 𝑦 2 − 17𝑦 + 70 = 0
Q263. I. 2𝑥 2 + 7𝑥 + 6 = 0
II. 3𝑦 2 + 11𝑦 + 10 = 0
Q264. I. 𝑥 2 − 2𝑥 − 24 = 0
II. 𝑦 2 − 12𝑦 + 36 = 0
Q278.
I. 𝑥 2 − 15𝑥 + 54 = 0
II. 𝑦 2 + 10𝑦 − 96 = 0
Q265. I. 4𝑥 2 + 11𝑥 + 6 = 0
II. 𝑦 2 + 10𝑦 + 25 = 0
Q279.
I. x3+440=2168
II.y2−23=121
Q266. I. 4𝑥 2 − 20𝑥 + 25 = 0
II. 5𝑦 2 − 6𝑦 − 8 = 0
32
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Directions (276-290): In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(a) if x>y
(b) if x≥y
(c) if x<y
(d) if x ≤y
(e) if x = y or no relation can be established between x and
y.
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Directions (291-295): Pie chart given below shows
distribution of passenger travelling from Haryana
roadways to different district. Read the data carefully and
answer the questions.
Q280. I. 𝑥 2 + 4𝑥 − 12 = 0
II. 𝑦 − 9𝑦 + 20 = 0
2
Q281. I. 𝑥 2 − 25𝑥 + 100 = 0
Total no. of passenger travelling
from Haryana roadways = 22500
II. 𝑦 2 − 27𝑦 + 110 = 0
10%
Q282. I. 𝑥 2 = 289
23%
II. 𝑦 = √289
12%
Q283. I. 𝑥 2 + 12𝑥 + 32 = 0
II 𝑦² + 7𝑦 + 12 =0
18%
22%
Q284. I. 3𝑥 + 16𝑥 + 20 = 0
2
II. 𝑦² + 14𝑦 + 48 =0
15%
Q285. I. 𝑥 2 + 𝑥 − 72 = 0
II. 𝑦 2 + 13𝑦 + 42=0
Gurgaon
Hisar
Sonipat
Panipat
Rewari
Ambala
Q291. No. of passenger who are travelling to Gurgaon are
approximately how much percent less than no. of
passenger travelling to Sonipat and Ambala together?
(a) 75%
(b) 78%
(c) 50%
(d) 65%
(e) 90%
Q286. I. 𝑥 2 + 5𝑥 + 6 = 0
II. 𝑦 2 − 9𝑦 + 14 = 0
Q287. I. 𝑥 2 − 14𝑥 + 45 = 0
II. 𝑦 2 + 2𝑦 − 35 = 0
Q288. I. 𝑥 2 + 11𝑥 + 18 = 0
Q292. What is the average no. of passengers who are
travelling to Hisar, Panipat and Rewari?
(a) 3025
(b) 2075
(c) 3375
(d) 3425
(e) 3075
II. 𝑦 2 + 6𝑦 + 8 = 0
Q289. I. 𝑥 2 + 5𝑥 = −6
II. 𝑦² – 15𝑦 = 16
Q290. I. 2𝑥 + 3𝑦 = 3
Q293. Passenger travelling to Hisar district are how many
less than passenger travelling to Ambala?
(a) 2525
(b) 2575
(c) 2425
(d) 2475
(e) None of these.
II. 3𝑥 + 𝑦 = 8
Q294. If ratio of men to women who are travelling to
Ambala and Gurgaon are 18:5 and 7:8 respectively, find
ratio between men travelling to Gurgaon and women
travelling to Ambala?
(a) 5:7
(b) 7:18
(c) 7:5
(d) 14:15
(e) 15:8
33
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Q295. If fair of a ticket for Rewari is Rs.75 and fair for
1
Panipat is 33 3 % more than that of Rewari, find difference
between total revenue generated from both district (in
Rs.)?
(a) 33750
(b) 22025
(c) 34250
(d) 35750
(e) 25075
Directions (296-300): Paragraph given below gives
information of literate and illiterate population out of total
population of three cities i.e. A, B and C. Read the paragraph
carefully and answer the following questions.
Total population of city A and B are 22000 and 16000
respectively. Total literate population of city B is 6000
which is 6.25% of total population of city C. Ratio of literate
to illiterate population in city A and C is 5:6 and 2:1
respectively. 40% of literate population in each city is
graduate.
Q296. Literate population from city B are what percent of
illiterate population of city A?
(a) 100%
(b) 75%
(c) 50%
(d) 40%
(e) 60%
Q297. What is the ratio between graduate population of
city C and total population of city B?
(a) 5:8
(b) 3:5
(c) 5:3
(d) 8:5
(e) 1:3
Q300. If ratio of male to female in graduate population
from city C is 9:7, find difference between graduate male
from city C to literate but ungraduated from city B?
(a) 7200
(b) 14400
(c) 10800
(d) 12000
(e) 11800
Directions (301-310): Bar graph given below shows
quantity of five different products (i.e. rice, pulse, wheat,
sugar and salt) sold (in kg) by a shopkeeper and table
shows total revenue (in Rs.) generated by selling these
individual products.
Total amount sold of each product (kg)
60
55
50
45
40
35
rice
pulses
wheat
sugar
Name of product
Total revenue (in Rs.)
Rice
2200
Pulse
3750
Wheat
900
Sugar
1200
Salt
600
salt
Q298. What is the difference between graduate population
of city B and illiterate population of city C?
(a) 29600
(b) 28400
(c) 28600
(d) 29400
(e) None of these.
Q301. Cost price of per kg rice is how much more or less
than per kg selling price of sugar when rice is sold at 60%
profit?
(a) Rs. 4 more
(b) Rs. 5 less
(c) None of these.
(d) Rs. 4 less
(e) Rs. 5 more
Q299. Population which is literate but ungraduated from
city A are what percent graduate population of city B?
(a) 500%
(b) 250%
(c) 300%
(d) 120%
(e) 375%
Q302. If 3 kg of wheat and 2 kg of salt is mixed, then what
will be the selling price per kg of such mixture?
(a) Rs.15
(b) Rs. 17
(c) Rs. 14
(d) Rs. 12
(e) Rs. 16
34
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Q303. Total revenue generated from wheat is what
percent of difference between total revenue generated
from rice and salt?
(a) 40.25%
(b) 56.25%
(c) 64.25%
(d) 45.50%
(e) 25.75%
Q304. If cost price of per kg pulse is Rs. 60, find profit
earned on selling 40 kg of pulse (in Rs.)?
(a) 450
(b) 600
(c) 800
(d) 750
(e) 300
Q305. What is the average quantity of rice, pulse and
wheat sold by shopkeeper?
(a) 45 kg
(b) 55 kg
(c) 60 kg
(d) 40 kg
(e) 50 kg
Directions (306-310): Study the following bar graph and
answer the questions that follow.
Given below is the bar graph which shows the number of
students playing three different games in five colleges in
year 2014.
NOTE- one student plays only one sport
Football
Cricket
Hockey
2
Q307. If 14 7 % of student playing Cricket of college N left
playing cricket and started playing Football in same college
then find the ratio of number of student playing football of
college N and M together to the number of student playing
Cricket of college K and N together?
(a) 3 : 2
(b) 1 : 2
(c) 1 : 1
(d) 1 : 3
(e) 2 : 1
Q308. Average no. of students playing Hockey of college K,
L and O is how much more than average number of
students playing football of college K, L & M ?
(a) 120
(b) 50
(c) 80
(d) 40
(e) 100
Q309. Total number of student playing Cricket of college L
and M together are what percent more/less than total
number of student playing Hockey of college K and M
together?
1
(a) 32 3 %
(b) 17
9
13
3
%
600
(c) 12 13 %
550
(d) 23 %
2
3
500
(e) 7
450
400
350
300
250
200
K
L
M
N
O
1
Q306. If 11 % of students playing Hockey of college L are
9
females then, number of males playing Hockey from same
college is what percent of average number of students
playing Hockey from college M & O ?
8
(a) 88 %
9
1
9
13
%
Q310. If total number of students in college K in year 2015
is increased by 20% percent with respect to year 2014 and
the ratio of student playing Football, Cricket and Hockey
becomes 5 : 2 : 3 respectively then find the average number
of students playing football in same college K in year 2014
and 2015 ?
(a) 640
(b) 525
(c) 625
(d) 545
(e) 454
(d) 72 7 %
Direction (311-315): Given bar graph shows total
number of confirmed cases of COVIND-19 and number of
deaths in four different countries. Study the bar graph
carefully and answer the questions given below.
(e) 82 3 %
Mortality rate = 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒐𝒕𝒂𝒍 𝒄𝒐𝒏𝒇𝒊𝒓𝒎𝒆𝒅 𝒄𝒂𝒔𝒆𝒔 × 𝟏𝟎𝟎
(b) 63 %
3
8
(c) 68 9 %
2
2
35
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒅𝒆𝒂𝒕𝒉
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400000
Q315. If number of confirmed cases in China is increased
by 25% and mortality rate remains same, what will be the
new number of total deaths in China.
(a) 4400
(b) 4500
(c) 4600
(d) 5200
(e) 5000
350000
350000
300000
250000
200000
140000
130000
80000
100000
50000
15000
17500
11000
4000
Spain
Italy
USA
China
0
Total confirmed cases
Deaths
Q311. For which country mortality rate is lowest among
the given four countries.
(a) Italy
(b) USA
(c) Spain
(d) China
(e) USA and China
Q312. Total confirmed cases in USA is what percent more
than total deaths in Italy.
(a) 1200%
(b) 1350%
(c) 2100%
(d) 1900%
(e) 1500%
Q313. Find out the ratio between mortality rate of Spain to
that of China?
(a) 19: 11
(b) 43:14
(c) 15:7
(d) 14:9
(e) 13: 5
Q314. Total death in all four countries together is what
percent of total confirmed cases in China?
(a) 59.375%
(b) 62%
(c) 55%
(d) 66.66%
(e) 75%
36
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Direction (316 – 320): Given below the bar graph shows
the quantity of six different items (in kg) purchased by a
person during the lockdown period. Read the data
carefully and answer the questions.
35
Items purchased (in kg)
150000
30
30
25
25
20
20
18
15
15
12
10
Sugar
Tea
Salt
Rice
Oil
Pulse
Q316. If the sum of the price of one kg sugar and one kg
salt is Rs.84 and the ratio of price of one kg of sugar and
one kg of salt is 11: 10. Then, find the difference between
the total price of Sugar and salt purchased by man?
(a) Rs. 220
(b) Rs. 240
(c) Rs. 260
(d) Rs. 300
(e) Rs. 280
Q317. If the total price of tea is Rs. 900 and that of rice is
Rs. 1500, then find the price of one kg tea is what percent
more than that of rice?
(a) 0%
(b) 20%
(c) 5%
(d) 10%
(e) 15%
Q318. If the price of one kg of pulse and one kg of oil is Rs.
63 and Rs. 42 respectively, then find the ratio of the total
price of the pulse to the total price of oil?
(a) 13:25
(b) 1:2
(c) 3:5
(d) 18:25
(e) 12:13
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Q319. The total quantity of sugar and salt purchased
together by man is what percent of the total quantity of rice
and pulse together purchased by man?
1
(a) 87 3 %
1
(b) 83 3 %
(c) 74%
(d) 92%
Q324. Deepak takes 24 minutes more to cover a certain
distance by decreasing his speed by 25%. What is the time
taken by him to cover the distance with his original speed?
(a) 70 minutes
(b) 72 minutes
(c) 75 minutes
(d) 90 minutes
(e) 84 minutes
1
(e) 64 3 %
Q320. If the price of one kg salt, one kg rice, and one kg oil
is Rs. 56, Rs. 32 and Rs. 40 respectively, then find out the
total price of oil, salt, and rice purchased by man?
(a) Rs. 2000
(b) Rs. 2800
(c) Rs. 2200
(d) Rs. 1800
(e) Rs. 2600
Q321. Train A crosses a 230m long platform in 29 seconds
and train B crosses a 150m long platform in 24 seconds.
Train B which is 450m long crosses train A in 160 seconds,
while running in the same direction. Find how much time
will the train A take to cross a 50m long bridge?
(a) 16 seconds
(b) 22 seconds
(c) 20 seconds
(d) 17 seconds
(e) 25 seconds
Q325. The speed of boat in downstream is ‘X-4’ kmph and
ratio of time taken by a boat to cover a certain distance in
upstream to downstream is 2 : 1. If boat takes 5 hours to
Cover 40 km in Upstream, then find the value of X?
(a) 16
(b) 20
(c) 22
(d) 24
(e) 18
Q326. Distance between two cities P and Q is 900 km. Car
A and Car B can cover the distance between P and Q in ‘X’
hours and (X + 4) hours respectively. If Car B and Car A
start from city P at 6.00 am and 8.00 am respectively and
both Cars meet at 10.30 am, then find the distance between
P and the point where both the cars meet?
(a) 425 km
(b) 475 km
(c) 450 km
(d) 500 km
(e) 400 km
Q322. A 950 metres long train-A crosses another train-B
running in same direction in 16 seconds. If the ratio of
speed of these trains is in the ratio 17:13 respectively, find
out the length of train B?
(a) 1000 meter
(b) 1900 meter
(c) 1600 meter
(d) 1100 meter
(e) Can’t be determine
Q327. Downstream speed of a boat is 33 3 % more than its
Q323. A train crosses a tunnel which is half of its length
with a speed of 144 km/hr. in ½ min, then find the time in
which it will cross another train which is double of its
length and standing on platform in opposite direction with
60% of its initial speed ?
(a) 120 sec.
(b) 90 sec.
(c) 150 sec.
(d) 100 sec.
(e) 180 sec.
Q328. Amit goes to office from his home by bike at the
speed of 30 kmph and he comes back to his home from
office by bike at the speed of X kmph. If average speed for
whole journey is 33 kmph, then find the value of ‘X’
(nearest to two decimal places)?
(a) 35.56 km/hr
(b) 36.00 km/hr
(c) 36.67 km/hr
(d) 32.50 km/hr
(e) 34.50 km/hr
37
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1
upstream speed and the speed of the boat in still water is
15 km/h more than the speed of the stream. Find the total
time taken by boat to travel 120 km in upstream?
(a) 7 hr
(b) 8 hr
(c) 9 hr
(d) 5 hr
(e) 10 hr
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Q329. A train ‘X’ starts from station P at 8 am and reaches
station Q at 4 pm. Another train ‘Y’ started from Q at the
same time at which ‘X’ started and reaches ‘P’ at 3 pm. then
find the time at which both the trains crossed each other.
(a) 11 : 44 am
(b) 11 : 48 am
(c) 11 : 36 am
(d) 12 : 44 pm
(e) 11 : 50 am
Q330. A car covered a certain distance at a certain speed
in a fixed time. If car had moved 9 kmph slower, it would
have taken 2 hours more and if it had moved 5 kmph faster,
it would have taken 48 min less. Find the distance covered
by car?
(a) 300 km
(b) 360 km
(c) 320 km
(d) 400 km
(e) 450 km
1
Q331. Downstream speed of a boat is 57 % more than the
7
upstream speed of a boat. If the speed of the stream is 8
km/hr., then find the total time taken by the boat to cover
176 km in downstream and 70 km in upstream.
(a) 7 hours
(b) 6.5 hours
(c) 7.5 hours
(d) 6 hours
(e) 8 hours
Q332. Speed of a boat in still water is 8 km/h. It takes 5
hours to go upstream and 3 hours downstream distance
between two points. What is the speed of stream?
(a) 4 km/h
(b) 2 km/h
(c) 3 km/h
(d) 1 km/h
(e) 2.5 km/h
Q334. A man can row 12 kmph in still water and it takes
him 90 minutes to reach a place & return. If the speed of
current is 4 kmph then how far is the place?
(a) 8 km
(b) 6 km
(c) 10 km
(d) 12 km
(e) 16 km
Q335. A man travels some journey on car with speed 60
kmph and some on cycle with speed 4 kmph. In return
journey he come in train with speed 20 kmph and take
equal time in both side journey. Find the ratio of the
distance travel by car, cycle and train.
(a) 8 : 2 : 11
(b) 3 : 2 : 5
(c) 2 : 1 : 3
(d) 6 : 1 : 7
(e) None of these
Q336. A spherical ball of radius 16 cm is melted and casted
into two cones of equal size and shape. If the base radius of
the cone is 50% of the height of the cone. Find the height of
each cone?
(a) 36 cm
(b) 18 cm
(c) 32 cm
(d) 20 cm
(e) 16 cm
Q337. How many three letters words starting with S (with
or without meaning) can be formed out of the letters of the
word, "STRANGE", if repetition of letters is not allowed?
(a) 10
(b) 15
(c) 12
(d) 30
(e) 18
Q333. A man covers half of total distance with 12 km/h
and another half distance with 24km/h. Find his
average speed.
(a) 12 km/h
(b) 16 km/h
(c) 10 km/h
(d) 18 km/h
(e) 6 km/h
38
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Q338. If two dice are rolled simultaneously, find the
probability of obtaining the sum (of numbers on these two
dices) which is divisible by 2 or 3 but not by both?
1
(a)
4
1
(b) 2
(c)
(d)
1
5
1
6
1
(e) 3
Q339. The area of a rectangular field having length 128m
and breadth 16m is equal to the area of an isosceles right1
angle triangle. If the radius of a sphere is 12 % of the
2
hypotenuse of the isosceles right-angle triangle, then find
out the total surface area of sphere?
(a) 512π m2
(b) 343π m2
(c) 580π m2
(d) 494π m2
(e) 500π m2
Q340. Gurdeep Chhabra joined ‘Adda 247’ with the work
experience of 26 years due to which average work
experience of all employees of ‘Adda 247’ was increased by
one year. If initial average work experience of all
employees of ‘Adda 247’ was five years, then find the new
number of employees in ‘Adda 247’?
(a) 23
(b) 19
(c) 25
(d) 21
(e) 27
Q341. If two dices are rolled together, then find the
probability of getting a number of one dice greater than the
number on other dice?
3
(a) 4
2
(b) 3
(e)
Q344. How many Words can be formed from the letters of
the word ‘FLAGSHIP’ so that the vowels always come
together?
(a) 5040
(b) 10080
(c) 720
(d) 360
(e) 1440
Q345. One card is picked randomly from a pack of 52
playing cards. What is the probability that it would either
be black queen or red king?
1
(a) 13
(b)
(c)
5
13
6
13
7
(d) 13
(e)
8
13
Q346. The ratio of height of a cylinder to its base radius is
2:1 respectively. If radius of a hemisphere is equal to the
radius of the cylinder, then find the total surface area of
cylinder is what percent more than total surface area of a
hemisphere?
(a) 40%
(b) 30%
(c) can’t be determined
1
1
(d) 33 3 %
5
(e) 50%
(c) 6
(d)
Q343. How many cubes of 7.5 cm edge can be cut out from
a cube of 45 cm edge?
(a) 108
(b) 72
(c) 216
(d) 230
(e) 256
6
1
2
Q342. The radius of a cylinder & a sphere is same, and ratio
of height and radius of cylinder is 2 : 1.If the volume of
sphere is 288 π cm³ then find the volume of cylinder? (in
cm³)
(a) 438 π
(b) 426 π
(c) 420 π
(d) 432 π
(e) 444 π
39
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Q347. A bag contains 4 red, 3 orange and 2 green color
balls. Find the probability of selecting two same color balls
from the bag?
1
(a) 2
(b)
(c)
7
18
4
5
5
(d) 13
5
(e) 18
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Q348. Find the probability of eight letters word that can be
formed from the letters of the word ‘BLASTING’ so that
vowels always come together.
1
(a) 4
(b)
(c)
2
5
1
3
10
(d) 21
5
(e) 14
Q349. The total surface area of a cylindrical vessel is 1232
cm2 and the height of vessel is 2 times more than the radius
of vessel. Find the volume of cylindrical vessel?
(a) 4312 cm3
(b) 3201 cm3
(c) 3234 cm3
(d) 3256 cm3
(e) 3333 cm3
Q350. There are 5 red balls, 6 black balls and some green
colored balls in a box. If the probability of choosing a black
1
ball from the box is 3, then find the number of greencolored ball in the box?
(a) 5
(b) 4
(c) 6
(d) 8
(e) 7
Directions (351-355): Line graph below shows the
number of complaints received by four different network
operators (A, B, C & D) on two different days Tuesday &
Wednesday. Study the line graph carefully and answer the
following questions.
140
120
100
80
60
Q351. Find the difference between the total number of
complaints received on both days by all network
operators?
(a) 40
(b) 50
(c) 60
(d) 70
(e) 80
Q352. Total number of complaints received by C & D
together on Tuesday are what percent more/less than the
number of complaints received by A & B together on
Wednesday?
(a) 62.50%
(b) 63.63%
(c) 66.66%
(d) 33.33%
(e)11.11%
Q353. Find the ratio of number of complaints received by
B on both days to number of complaints received by A & D
together on Wednesday?
(a) 1: 2
(b) 2 : 1
(c) 1 :1
(d) 5 : 4
(e) 4 : 5
Q354. Find the total number of complaints received by C
on Tuesday and Wednesday are approximately what
percent of total number of complaints received on Tuesday
by all network operators together?
(a) 36%
(b) 60%
(c) 53%
(d) 48%
(e) 67%
Q355. Find the ratio of complaints received by A,B & D
together on Tuesday to total complaints received on
Wednesday by all network operators together?
(a) 11: 17
(b) 21 : 31
(c) 18 : 19
(d) 29 : 32
(e) 51 : 43
Direction (356–360): In each of these questions a number
series is given. In each series only one number is wrong.
Find out the wrong number.
40
20
0
A
B
Tuesday
40
C
D
Wednesday
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Q356. 8, 4, 4, 10, 12, 30, 90
(a) 90
(b) 8
(c) 10
(d) 12
(e) 30
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Q357. 11, 16, 25,
(a) 41
(b) 66
(c) 11
(d) 151
(e) 25
41, 66,
102, 151
Q364. 3167 − 2881 − 112 =? −√1681
(a) 316
(b) 416
(c) 286
(d) 326
(e) 206
Q365. 62.5% 𝑜𝑓 ? −(5)2 = 152
(a) 200
(b) 100
(c) 500
(d) 400
(e) 300
Q358. 21, 25, 20, 28, 19, 27, 18
(a) 18
(b) 27
(c) 19
(d) 25
(e) 20
Direction (366–370): What approximate value should
come in the place of question (?) mark in the following
questions.
Q359. 20, 28, 40, 56, 76, 104, 128
(a) 104
(b) 128
(c) 56
(d) 28
(e) 40
3
Q366. 24.01% of 449.98 + ?2 = (16.01)2 – √63.93
(a) 8
(b) 12
(c) 10
(d) 9
(e) 14
Q360. 1, 2, 6, 20, 88, 445, 2676
(a) 2
(b) 6
(c) 88
(d) 2676
(e) 20
Direction (361–365): What will come in the place of
question (?) mark in following the question:
Q361. 36 ÷ 4 × 7 + 4 × 4.5 =?2
(a) 9
(b) 7
(c) 19
(d) 17
(e) 3
Q368. 4? + 79.98% of 980.03 = 1039.99
(a) 4
(b) 2
(c) 3
(d) 5
(e) None of these
Q362. √1849 − √256 = √?− √144
(a) 1681
(b) 1600
(c) 1296
(d) 1446
(e) 1521
Q369.
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1512.01
(a) 64
(b) 32
(c) 48
(d) 36
(e) 42
Q363. 250% 𝑜𝑓 30 − 175% 𝑜𝑓 36 + 52 =?
(a) 27
(b) 18
(c) 37
(d) 21
(e) 31
41
Q367. ? × (44.01 % of 750.01 + 110.01) = 87.99% of
2499.98
(a) 2
(b) 4
(c) 3
(d) 5
(e) 6
?
+ 49.99% of 488 = 70.03% of 399.99
Q370. ?% of 639.98 + 40.03% of 279.99 = (19.99) 2
(a) 25
(b) 50
(c) 35
(d) 45
(e) 40
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Direction (371-375): Read the given information
carefully and answer the following questions.
Line graph shows production of three products in terms of
percentage (out of total production in the year) in four
different years.
Food Products
Dairy products
Beverages
70
Q374. The difference between number of food products
and dairy products produced in 2015 and 2018 together is
12000. Find the average of dairy products and beverages
produced by company in 2017?
(a) 30000
(b) 22500
(c) 20000
(d) 24000
(e) 25000
Q375. Find the total production in 2019 if there was an
increase of 20% in production in 2019 as compared to
previous year given that number of dairy products in 2015
was 18000?
(a) 1,20,000
(b) 1,08,000
(c) 1,18,000
(d) 1,12,000
(e) None of these
60
50
40
30
20
10
0
2015
2016
2017
2018
1.Company produces three different products i.e. food,
dairy and beverages.
2.Total production of the company was same in all years.
Q371. In 2016, quantity of food products and dairy
products produced is what percent more or less than that
of beverages produced in year 2015 and 2016 together?
1
(a) 13 %
3
(b) 15%
2
(c) 16 3 %
(d) 12.5%
(e) 10%
Q372. If total number of products produced in year 2018
was 1,50,000. Find the difference between number of food
products produced in 2017 and number of dairy products
produced in 2015 and 2016 together?
(a) 12000
(b) 18000
(c) 12500
(d) 10000
(e) 15000
Directions (376-380): What comes at the place of
question marks:
Q376. 588,
(a) 552
(b) 510
(c) 542
(d) 532
(e) 572
562 ,
614,
536,
640,
Q377. 27,
(a) 912
(b) 892
(c) 922
(d) 932
(e) 802
52,
102 ,
202,
402, ?
Q378. 17,
(a) 461
(b) 481
(c) 471
(d) 491
(e) 451
41,
91 ,
171,
293,
?
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?
Q373. Find the ratio of average of number of food products
produced in 2015, 2017 and 2018 to total number of
beverages produced in 2016 and 2017 together.
(a) 3: 2
(b) 2: 3
(c) 3: 5
(d) 5: 3
(e) 3: 4
42
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Q379. 35,
(a) 9.62
(b) 8.76
(c) 12.56
(d) 10.08
(e) 11.02
7,
42,
8.4,
Q380. 24, 60 ,
(a) 812.75
(b) 843.75
(c) 792.75
(d) 875.75
(e) 896.75
90,
225,
50.4,
Q383. Find out the ratio between selling price of
refrigerator C in 2018 and the selling price of refrigerator
A in 2017?
(a) 20:21
(b) 18:25
(c) 19:25
(d) 23:27
(e) 16:25
?
337.5,
?
Direction (381-385): Line graph given below shows the
selling prices (in rupees) of three types of Refrigerators (A,
B & C) in four different years i.e. 2016, 2017, 2018, and
2019 for a shopkeeper
35000
30000
25000
20000
15000
10000
2016
2017
Refrigerators A
2018
2019
Refrigerators B
Refrigerators C
Q381. If a discount of 24% is given on refrigerator C sold
in 2018 and ratio of MP to CP of C in 2018 is 5 : 3. then find
the difference between the discount allowed and profit
earned on C in 2018. (in rupees)
(a) 5000
(b) 4000
(c) 2000
(d) 6000
(e) 3000
Q382. Find out average selling price of refrigerator A in all
the given years. (in rupees)
(a) 16500
(b) 22500
(c) 18500
(d) 19500
(e) 25500
43
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Q384. In which year sum of selling price of all 3 type of the
refrigerator was the lowest?
(a) 2019
(b) 2016
(c) 2017
(d) 2018
(e) 2016 and 2018
Q385. selling price of refrigerator A in year 2018 is approx.
what percent of selling price of refrigerator B in 2019?
(a) 78%
(b) 88%
(c) 82%
(d) 72%
(e) 93%
Direction (386–390): In each of these questions, two
equations (I) and (II) are given. You have to solve both the
equations and give the answers accordingly.
(a) if x>y
(b) if x≥y
(c) if x<y
(d) if x ≤y
(e) if x = y or no relation can be established between x and
y.
Q386. I. x² – 14x + 48 = 0
II. y² – 17y + 72 =0
Q387. I. x2 + 13x + 42 = 0
II . y2 + 15y + 56 = 0
Q388. I . x2 + 8x + 12 = 0
II . 6y2 + 13y + 6 = 0
Q389. I. 2x2 + 9x + 9 = 0
II. y2 + 28y + 192 = 0
Q390. I . x² – 9x + 20 = 0
II. y² – 6y + 9 = 0
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Directions (391-395): What should come in place of the
question mark (?) in the following number series.
Q391. 90,
(a) 285
(b) 325
(c) 470
(d) 855
(e) 270
55,
75,
142.5,
? , 862.5
Q392. 5,
(a) 850
(b) 750
(c) 800
(d) 805
(e) 820
12,
39,
160,
4836
Q393. 26,
(a) 146
(b) 133
(c) 201
(d) 134
(e) 156
36, 54, 80, 114, ?
?,
Q397. If the total price of tea is Rs. 900 and that of rice is
Rs. 1500, then find the price of one kg tea is what percent
more than that of rice?
(a) 0%
(b) 20%
(c) 5%
(d) 10%
(e) 15%
Q398. If the price of one kg of pulse and one kg of oil is Rs.
63 and Rs. 42 respectively, then find the ratio of the total
price of the pulse to the total price of oil?
(a) 13:25
(b) 1:2
(c) 3:5
(d) 18:25
(e) 12:13
Q394. 17, 25, 49, 97, 177, ?
(a) 297
(b) 247
(c) 358
(d) 292
(e) 279
Q395. 21, 28, 42, 64, 95, ?
(a) 125
(b) 158
(c) 142
(d) 136
(e) 164
Q399. The total quantity of sugar and salt purchased
together by man is what percent of the total quantity of rice
and pulse together purchased by man?
1
(a) 87 3 %
Items purchased (in kg)
Direction (396–400): Given below the bar graph shows
the quantity of six different items (in kg) purchased by a
person during the lockdown period. Read the data
carefully and answer the questions.
35
30
25
20
15
10
Sugar
44
Tea
Salt
Q396. If the sum of the price of one kg sugar and one kg
salt together is Rs.84 and the ratio of price of one kg of
sugar and one kg of salt is 11: 10. Then, find the difference
between the total price of Sugar and salt purchased by
man?
(a) Rs. 220
(b) Rs. 240
(c) Rs. 260
(d) Rs. 300
(e) Rs. 280
Rice
Oil
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Pulse
|
1
(b) 83 3 %
(c) 74%
(d) 92%
1
(e) 64 3 %
Q400. If the price of one kg salt, one kg rice, and one kg oil
is Rs. 56, Rs. 32 and Rs. 40 respectively, then find out the
total price of oil, salt, and rice purchased by man?
(a) Rs. 2000
(b) Rs. 2800
(c) Rs. 2200
(d) Rs. 1800
(e) Rs. 2600
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Solutions
S4. Ans.(d)
20
Sol. Total number of students in A = 200 ×
= 40
S1. Ans.(d)
20
Sol. Total students in A = 200 × 100 = 40
100
22.5
Total students in C = 200 × 100 = 45
Difference between girls and boys in A = 50 ×
20
100
10
= 10
Difference between girls and boys in C = 50 × 100 = 5
Let total number of girls in A = x
So, total number of boys in A = (40 – x)
And let total number of girls in C = y
Total number of boys in C = (45 – y)
Given, (40 – x) − 𝑥 = 10
2x = 30
x = 15
Similarly, (45 – y) – y = 5
2y = 40
y = 20
25
25 25
Required probability = 40 × 45 = 72
( 25−𝑎 𝐶2 + 15−2𝑎 𝐶2 )
40−3𝑎 𝐶
2
=
11
18
a=4
Required difference = 2a – a = a = 4
S2. Ans.(b)
Sol. All five students from each B & D chosen for representing
their respective school in a debate competition should be boys
and number of girls should be greater than number of boys in
order to maximize the probability of girls in remaining
students.
15
Total number of students in B = 200 × 100 = 30
25
Total number of students in D = 200 × 100 = 50
20
Difference between boys and girls in B = 50 × 100 = 10
20
Difference between boys and girls in D = 50 × 100 = 10
Let total number of boys in B = a
So, total number of girls in B = (30 – a)
Let total number of boys in D = b
So, total number of girls in D = (50 – b)
ATQ,
(30 – a) – a = 10
a = 10
Also, (50 – b) – b = 10
b = 20
20
30
8
Required probability = 25 × 45 = 15
100
S5. Ans.(a)
Sol. Let total saving of Ritu be x Rs.
5𝑥
Invested in Scheme A = 15
𝑥
= 3 Rs.
4𝑥
Amount Invested in scheme B = 15 Rs.
2𝑥
Invested in scheme C = 5 𝑅𝑠.
10×10
Two year C.I on 10% =10+10+
100
= 21%
15×15
Two year CI on 15% =15+15 + 100
= 32.25 %
20×20
Two year CI on 20% =20+20+ 100
= 44%
ATQ,
4𝑥
32.25
𝑥
21
× 100 − 3 × 100 = 744
15
129𝑥
21𝑥
− 300 = 744
24x = 744 × 1500
744×1500
x = 24
x = 46500 Rs.
Required difference
46500×6
44
46500×4
32.25
=
×
−
×
1500
15
100
= 8184 – 3999
= 4185 Rs.
S3. Ans.(a)
17.5
Sol. Total students in E = 200 ×
= 35
30
Difference between girls and boys in E = 50 × 100 = 15
Let number of girls in E = p
So, number of boys in E = (35 – p)
Given, (35 – p) – p = 15
p = 10
Number of boys in F = (35 – 10) – 5 = 20
And number of girls in F = 10
20×10 40
Required probability = 30𝐶 = 87
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15
100
S6. Ans.(d)
Sol. Let total students participated in exam P & Q be ‘x’ & ’y’
respectively.
ATQ –
40
75
x × 100 × 100 = 900
3𝑥
= 900
x = 3000
40
(100−36)
Also, y ×
×
= 640
100
100
y = 2500
2500
Required ratio =
=5:6
10
3000
2
45
20
Difference between girls and boys in A = 50 × 100 = 10
Let total number of girls in A = x
So, total number of boys in A = (40 – x)
ATQ,
(40 – x) − 𝑥 = 10
2x = 30
x = 15
Let total number of boys who left school A = a
And let, total number of girls left school A = 2a
Now,
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S7. Ans.(e)
Sol. Let total students participated in exam from Q = a
ATQ –
60
60
a×
×
= 1440
100
100
a = 4000
40
40
Total girls failed from Q = 4000 × 100 × 100 = 640
100
100
= 90 km/hr
Car A speed in km/hr
750
18
=
×
100
60
S8. Ans.(a)
Sol. Let total number of students participated from S = p
ATQ –
45
50
p×
×
= 1125
100
100
p = 5000
Given, ratio of total failed boys to total failed girls = 7 : 3
55
7
3
Required difference = 5000 × 100 × (10 − 10) = 1100
S9. Ans.(b)
Sol. Let total students participated in exam from P = x
40
75
25
x × 100 × (100 − 100) = 600
𝑥
5
= 600
x = 3000
Let total students participated in exam from S = y
45
50
y × 100 × 100 = 1350
300
3300 −3000
3000
S13. Ans.(d)
1800
18
Sol. Speed of car B = 60 × 5 = 108 km⁄hr
1500
18
Speed of car C = 60 × 5 = 90 km⁄hr
Average speed of Rajeev whole journey
2xy
2×108×90
= x+y = 108+90
19440
Required distance = speed × time
19440
= 198 × 11 = 1080 km
Distance between city and village
1080
=
= 540 km
2
× 100
= 3000 × 100 = 10%
S10. Ans.(c)
Sol. Let total students participated from P be ‘2x’
So, total students participated from R = 3x
55
(100−30)
40
(100−25)
3x × 100 × 100 – 2x × 100 × 100 = 2220
1.155x – 0.60x = 2220 ⇒ x = 4000
40
25
Total girls passed from P & R = 8000 × 100 × 100 + 12000 ×
55
5
= 45 km/hr
Let both Car has travelled for t hours before they meet.
According to question
90t – 45t = 180
180
t= 45 = 4 hours.
Therefore, total distance traveled is (90 × 4) + (45 × 4) =
540𝑘𝑚.
= 198 km⁄hr
y = 6000
55
Total failed students from S = 6000 × 100 = 3300
Required percentage =
1080
D= 5 𝑘𝑚
D=216KM
2D = 432 km
18
b = 2000
45
42
Total girls failed from R = 2000 ×
×
= 378
100
D
+ 2 = 180
= 25 × 5
100
Required difference = 640 – 378 = 262
3
S12. Ans.(a)
Sol. Car C speed in km/hr
1500
18
= 60 × 5
Let total students participated in exam from R = b
So,
45
58
55
30
b×
×
−𝑏×
×
= 192
100
D
30
× 100
100
= 800 + 1980 = 2780
2780
Required average = 2 = 1390
S14. Ans.(c)
1200
18
Sol. Car E usual speed = 60 × 5
= 72 km⁄hr
Time taken from this speed is 480 minutes, i.e. 8 hours.
Hence total distance b/w City X & Y is 8 × 72 = 576 𝑘𝑚
Reduce speed = 72 – 12 = 60 km/hr
576
48
Required time = 60 = 5 hours or 576 minutes
S15. Ans.(a)
Sol. Time = t =time after which they meet
1200
18
Speed of car E = 60 × 5 = 72 km⁄hr
Speed of car C =
1500
60
×
18
5
= 90 km⁄hr
2
= 72 (t – 3) + 90t = 762
S11. Ans.(b)
Sol. Usual speed of car D in km/hr
900
18
= 60 × 5
=
15×18
5
= 72t – 48 + 90t = 762
762+48
=t=
162
= 54 km⁄hr
Let total distance b/w Delhi and Lucknow is 2D km`
According to question
D
D
+ = 10
54
t = 5 hr
Both cars meet at = 3.20 + 5 = 8.20 pm
2
Distance from Delhi = (5 – 3) 72
= 312 km
36
46
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S16. Ans.(e)
Sol. Pattern – 16 multiplied by consecutive prime numbers
16 × 2 = 32
16 × 3= 48
16 × 5 = 80
16 × 7 = 112
16 × 11 = 176
16 × 13 = 208
16 × 17 = 272
So, series is right no need to replace any number
S17. Ans.(b)
Sol. Wrong number = 8332
Pattern of series →
12 × 6 – 2 = 70
70 × 5 – 2 = 348
348 × 4 – 2 = 1390
1390 × 3 – 2 = 4168
4168 × 2 – 2 = 8334
8334 × 1 – 2 = 8332
So, 8332 should be replaced with 8334
S18. Ans.(c)
Sol. Pattern of series I –
2.4 ,
7.2 ,
4,
12 , 18.4 , 26.4 ,
+3.2 +4.8
+1.6
+1.6 +1.6
+6.4
+1.6
+9.6
+8.0
+1.6
36
+1.6
Same series II –
(A=18.4), 20 ,
+ 1 .6
28 , (B=34.4), 42.4
23.2 ,
+ 3 .2
+ 4 .8
+ 1 .6 + 1 .6
+ 6 .4
+ 1 .6
+ 9 .6
+ 8 .0
+ 1 .6
+ 1 .6
S19. Ans.(e)
Sol. Pattern of series I –
123 = (112 +2)
171 = (132 + 2)
227 = (152 + 2)
291 = (172 + 2)
363 = (192 + 2)
443 = (212 + 2)
531 = (232 + 2)
Same series II –
402 = (202 + 2)
So, (A) = (162 + 2) = 258
(B) = (182 + 2) = 326
(C) = (242 + 2) = 578
47
(C=52)
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S20. Ans.(a)
Sol. Pattern of series I –
Pattern of series I-
Same series II –
(A) = 1600 – 1280 = 320
(B) = 2400 + 560 = 2960
(C) = 3780 + 350 = 4130
S21. Ans.(c)
Sol. Amar attempted total of 75 question in which 9 do not
have negative marking.
Total correct questions of Amar
80
= 100 × 75 = 60
If his all non-negative marking questions were correct, then
his score
= 60 × 3 – (15 × 1) – (0.5 × 25)
[12.5 marks are subtracted for his un attempted questions]
= 180 – 27.5 = 152.5
If his all non-negative marking questions were wrong, it
means 6 of his questions which carry negative marks were
wrong.
= 60 × 3 – (6 × 1) – (0.5 × 25)
= 180 – 18.5
= 161.5
S22. Ans.(d)
Sol. Total questions attempted by Prem
80
= 100 × 75
= 60
Total correct questions of Prem
90
= 100 × 60
= 54
His 6 questions were wrong and 3 were negative marking
questions.
Therefore his score, is
= 54 × 3 – (3 × 1) – (40 × 0.5)
= 162 – 3 – 20
= 139
S23. Ans.(b)
Sol. Prem, attempted total of 60 questions and 54 were
incorrect.
34 questions of section A were correct.
50
He attempted, 16 question [ × 32] of section C and 6 were
100
incorrect.
Score from section C = 10 × 3 – (6 × 1) – (16 × 0.5) = 16
If 6 questions from section C were wrong, it means all the
questions, which he attempted from section B were correct.
He attempted total of (60 – 34 – 16) = 10 question from
section B
Score from section B = 10 × 3 – 24 (0.5)
= 30 – 12 = 18
Required difference = |16 – 18| = 2
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S24. Ans.(a)
Sol. Correct question for Amar
80
=
× 75 = 60
S30. Ans.(c)
240
Sol. Cost price per kg of sugar = 15 = 16
4
Discount offered by wholesaler = 20 × 100 = 20%
100
Score of Amar = 60 × 3 – (15 × 1) – (25 × 0.5)
= 180 – 27.5
= 152.5
Score of Prem = 54 × 3 – (40 × 0.5)
= 162 – 20
= 142
Required difference = 10.5
Discount offered by retailer to customer =
= 357
Profit %=
× 100 = 48.75%
2
3
65
× 100 = 3120
Total females do not get loan from village P
2
35
10
= 7200 × 3 × 100 × 21 = 800
Total males do not get loan from village P
1
= 7200 × 3 − 800 = 1600
Total males applied for loan from village P
= 3120 + 1600 = 4720
Total number of males get loan from village T
80
20
8
= 9600 × 100 × 100 × 11 = 960
Total males do not get loan from village T
31.25
= 9600 × 100 − 960 = 2040
S32. Ans.(e)
125
72
Sol. Total females do not get loan from S = 10000 × 100 ×
= Rs. 1500
24
× 100 = 66.66%
100
16
× 27 = 1024
Total females do not get loan from Q
60
3
Required percentage =
= Rs. 270
To be in a situation of no loss –no gain, he must sell remaining
50% at Rs. 270.
270
Price per kg =
= Rs. 27
10
S29. Ans.(d)
70
Sol. Cost price of pulses for the retailer = 18 × 35 × 100 = Rs.
441
85
Cost price of 2 kgs of Almond = 2 × 40 × 100 = Rs. 68
Total CP = 441 + 68 = Rs. 509
304
1024 −720
720
× 100
2
= 720 × 100 = 42 9 %
S33. Ans.(d)
Sol. Total females do not get loan from Q
= 8000 ×
3
×
5
25
100
3
× = 720
5
40
Total males do not get loan from Q = 8000 × 100 − 720 = 2480
Total females do not get loan from S
72
24
16
= 10000 × 100 × 100 × 27 = 1024
Total males do not get loan from S
130
Total SP = 18 × 35 × 100 = Rs. 819
28
= 10000 × 100 − 1024 = 1776
310
2480
× 100 = 509 × 100 ≈ 61%
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25
= 8000 × 100 × 100 × 5 = 720
S28. Ans.(c)
90
Sol. Cost price of Rice for Retailer = 20 × 15 × 100
48
240
Total males applied for loan from village T
= 5280 + 2040 = 7320
Required difference = 7320 – 4720 = 2600
Price at which he sold all the cashew = 40 × 30 × 100
509
117
S31. Ans.(a)
Sol. Total number of males get loan from village P = 7200 ×
68.75
S27. Ans.(b)
Sol. Cost price of cashew for retailer = Rs. 900
819−509
× 100 =
Total females do not get loan from village T
= Rs. 475
Profit = 475 – 450 = 25
Profit % =
240
68.75
Price at which he sold this amount of wheat to customer
95
= 20 × 25 × 100
900
357−240
= 9600 × 100 × 100 = 5280
S26. Ans.(a)
Sol. Cost price of 20 kg of wheat for retailer
90
= 20 × 25 × 100 = Rs. 450
1500−900
× 20% = 15%
Selling price of mixture = (6 + 15) × 20 × 100
S25. Ans.(e)
Sol. Let he attempted x correct questions.
& (75 – x) wrong questions. But (75 – x – 4) carries negative
marking. Therefore
His score
3x – (71 – x) – 25 × (0.5) = 108.5
= 3x – 71 + x = 121
4x = 192
x = 48
hence option (e)
Profit % =
75
100
85
Required ratio = 1776 = 155 ∶ 111
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S34. Ans.(c)
Sol. Total number of males who get loan from P
2
65
= 7200 × ×
= 3120
3
S39. Ans.(d)
Sol. Wrong number = 19480
Pattern of series –
2.5 × 24 = 60
60 × 12 = 720
720 × 6 = 4320
4320 × 3 = 12960
12960 × 1.5 = 19440
19440 × 0.75 = 14580
So, there should be 19440 instead of 19480.
100
Total number of males who get loan from Q
60
3
= 8000 ×
× = 3600
100
4
Total number of males who get loan from R
= 8800 ×
9
11
×
82
100
= 5904
Total number of males who get loan from S
72
76
= 10000 × 100 × 100 = 5472
Total number of males who get loan from T
68.75
80
= 9600 × 100 × 100 = 5280
S40. Ans.(a)
Sol. Wrong number = 200
Number of males(second highest) who get loan are from
village S.
Total females applied for loan from village S
72
24
43
= 10000 × 100 × 100 × 27 = 2752
So, there should be199 instead of 200.
S35. Ans.(e)
Sol. Total females who get loan from village R
9
18
= 8800 × 11 × 100 = 1296
S41. Ans.(b)
Sol. If each works 2 days at a time alternately starting with A,
the work is completed in exactly 10 days.
∴ A works for 6 days and B worked for 4 days.
6
4
+b=1
………….(i)
a
Total females who do not get loan from village R
4
= 1296 × 9 = 576
Total males who do not get loan from village R
= 8800 ×
2
11
If B starts, the work is completed in 10.5 days.
∴ B works for 6 days and A worked for 4.5 days.
6
4.5
+ a =1
………….(ii)
b
− 576 = 1024
Required average =
1024+1296
2
= 1160
By solving (i) and (ii)
a = 9 days
And, b = 12 days
Time taken by A and B working together to complete the work
1
1
=1 1=1 1
S36. Ans.(c)
Sol. Wrong number = 558
+
+
a b
9 12
36
1
= 7 = 5 7 days
So, there should be 550 instead of 558.
S42. Ans.(a)
P
22P–36
1
⇒ P² – 204P + 1188 = 0
⇒ P² – 198P – 6P + 1188 = 0
⇒ P = 198 or 6
⇒P=6
Required number of balls in bag B = 48.
S38. Ans.(e)
Sol. Wrong number = 1260
S43. Ans.(c)
Sol. Let that distance be D km.
And, speed of boat P in still water be 2x km/hr
Speed of boat Q in still water be 3x km/hr
Speed of stream be r km/hr
ATQ 𝐷
3𝐷
= 5×(2𝑥−𝑟)
3𝑥−𝑟
So, there should be 1258 instead of 1260
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1
⇒ P2 +60P+756 = 12
So, there should be 5.5 instead of 4.5.
49
P+2
Sol. P+18 – 42+P = 12
S37. Ans.(a)
Sol. Wrong number = 4.5
Or, 𝑥 = 2𝑟 ……..(i)
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Also,
(3𝑥 + 𝑟) × 5 − 5 × (2𝑥 + 𝑟) = 60
or, 𝑥 = 12
speed of stream= 6km/hr
S47. Ans.(d)
Sol. Let, the income, expenditures and saving of P, Q and R:
P
Q
R
3
3
Income
x + 8000 x
x
4
4
Expenditure on Rent y
y
y
30
Required ratio=30 = 1: 1
Solutions (44-45)
Man → 12 × 6 days
Woman → 8 × 18 days
Child → 10 × 18 days
(4 men + 12 women + 20 children)’s 2 day work
4
= 2[
12×6
=
+
12
8×18
+
20
10×18
]
2
1
Remaining work = 1 − 2 = 2
According to question —
1
2
36
12×6
z
z + 1000
z + 2000
Savings
6t
7t
4t
Now,
Total amount spent by all the three on food = 27000
⟹ z + z + 1000 + z + 2000 = 27000
⟹ z = 8000
Monthly income of Q = Monthly income of P + 6000
3
⟹ x = x + 8000 + 6000
4
⟹ x = 56000
Savings of P
6
=
1
1
Expenditure on Food
Savings of Q
7
3
x + 8000 − y − z
4
6
⟹ x − y − z −1000 = 7
=𝑥
Putting the values of x and z
42000 − y
6
⟹ 47000 − y = 7
𝑥=1
⟹ y = 12000
Monthly rent of the apartment = 3y = Rs.36000
S44. Ans.(c)
Sol. A → 10 days
B → 20 days
S48. Ans.(e)
Sol. Let S.P. → 120x
So,
C.P. of 1st Article → 100x
C.P. of 2nd Article → 96x
Profit difference = 4x → 8 Rs.
x = 2 Rs.
Selling price = 240 Rs.
2 days’ work of (A + B) = 3
12 days’ work of (A + B) = 18
A does the remaining work (20 – 18 = 2 units)
2
Required time = 12 + 2 = 13 days.
S49. Ans.(d)
Sol. Interest earned from first scheme
S45. Ans.(a)
Sol. Soldier = 56 × 1 × 24 days
Required time =
56×24
42
R
𝑅
Rate of interest for second scheme = 𝑅 [1 + 100] =
Interest earned from second scheme
= 32 days
S46. Ans.(a)
Sol. Let, the income, expenditures and saving of P, Q and R:
P
Q
R
3
x + 8000
x
3
Expenditure on Rent
y
y
y
Expenditure on Food
z
z + 1000
z + 2000
Savings
6t
7t
4t
Income
4
4
100
…(ii)
S50. Ans.(e)
4 2
Sol. Probability of a hit by Sunny = 6 = 3
1
For a miss = 3
5
Savings of Q = 62 2 % of income of Q = 8 x
4
5
5
7
8
14
∴ Savings of R = × x =
1
For a miss = 2
x
1
6
2
2
1
1
2
1
1
1
1
2
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Required probability = + × × + × × × × +. ..
3
3
2
3
3
2
3
2
3
It formed a G.P.
2
1
1 1
In which a = and r = × =
Percent of R’s savings out of his monthly income
x
= 14
3 × 100
3
x
3
2
a
6
2
4
Required probability = 1–r = 31 = 5
13
= 47 21 %
50
3
Probability of a hit by Satish = =
Savings of Q and R are in the ratio 7 : 4.
4
x
15000×2×(100R+R2 )
=
= 300𝑅 + 3𝑅2
100×100
ATQ,
300𝑅 + 3𝑅 2 − 1.5𝑅2 − 300𝑅 = 600
1.5𝑅2 = 600
⇒ R = 20%
100𝑅+𝑅 2
1
Now,
5
2
= 15000 [(1 + 100) – 1] = 1.5𝑅2 + 300𝑅…(i)
1–
6
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S51. Ans.(d)
Sol. Probability of a girl being selected from a section =
Total girls in the section
Total students in the section
Let the number of girls, number of boys and total number of
students respectively:
For section A: 2x, 3x and 5x.
For section B: 4y, 5y and 9y.
For section C: 5z, 4z and 9z.
According to the question,
Ratio of total number of students in the three sections:
⟹ 5x : 9y : 9z = 10 : 12 : 9
⟹x:y:z=6:4:3
Let the values of x, y and z be 6k, 4k and 3k respectively.
Total number of girls in all the three sections
= 2x + 4y + 5z = 12k + 16k + 15k = 43k
Total number of students in all the three sections
= 5x + 9y + 9z = 30k + 36k + 27k = 93k
Probability of a girl being selected from the students from all
the three sections together
Total girls in all sections
43k
43
= Total students in all sections = 93k = 93
S52. Ans.(c)
Sol. According to the question,
Number of girls in sections A = Number of boys in section C
⟹ 2x = 4z
⟹ x = 2z
Number of boys in section A : Number of boys in section C
= 3x : 4z = 6z : 4z = 3 : 2
S53. Ans.(a)
Sol. Let total quantity of milk = 200x L
And total quantity of water = 100x L
Total milk in P and Q = (20% + 15%) 200x = 35 × 2x = 70x L
Total water in P and Q = 35 × x
25
25
Total water in X = 35x + 100 × 100 × 100x
= 35x + 6.25x
= 41.25x L
Let cost price of milk per liter be Rs.10
So, cost price of (70x + 41.25x) L of mixture
= 70x × 10 = Rs.700x
Selling price of (70x + 41.25x) L of mixture
= 111.25x × 10 = Rs.1112.5x
Required profit percentage =
=
1112.5x−700x
700x
S54. Ans.(c)
50
Sol. Milk in vessel A and C = 100 × 2x = x
Water in vessel A and C =
× 100
7
825
= 14
= 0.55x
Ratio of milk and water in M = x : 0.55x
= 20 : 11
According to question,
⇒
20
×62
31
11
55x− ×62+17
31
x−
x−40
6
=5
6
⇒ 55x−5 = 5
⇒ 5x – 200 = 3.30x – 30
x = 100
Quantity of milk in vessel B =
20
100
× 2 × 100
= 40 L
S55. Ans.(b)
Sol. Let total milk in all 5 vessel = 200x
And total water in all 5 vessel = 100x
So,
65
Total milk in all vessel except C = 100 × 200x
= 130x
55
Total water in all vessel except C = 100 × 100x
= 55x
And
Ratio of milk and water in vessel C = 35 × 2x : 45x
= 70x : 45x
= 14 : 9
According to question,
14
×115
23
9
55x+ ×115
23
130x+
130x+70
55x+45
9
=4
9
=4
520x + 280 = 495x + 405
25x = 125
x=5
Total quantity of water in all five vessel = 100x = 500 L
5x
3
C → 35 = 7 or x = 3
13
= 5814%
Or we can say that profit in due the quantity of water in the
mixture.
So we can directly write
41.25x
70x
× 100
= 5814%
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From A and C
Blue =15
Black=12
Red=8
So, required probability =
13
51
×x
S56. Ans.(c)
Sol. A → Let number of black and blue colored balls be 4x and
5x respectively.
B → Let number of red colored balls be y
So, y+4x=5x+5
412.5
% profit =
55
100
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S57. Ans.(e)
Sol. A → Let the length and breadth of the rectangle be 4x m
and 3x m respectively.
4𝑥+3𝑥
Side of that square=
= 3.5𝑥 𝑚
2
B → Sum of the lengths of diagonals of the rectangle =
(𝑆𝑖𝑑𝑒 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒) × 3 − 4
C → Area of a square = 784 𝑚2
Hence, the question can be answered by using any two of the
three statements together.
Hence, It can be solved either by A and B together or by A and
C together
S58. Ans.(a)
Sol. Let the ten’s place and unit’s place digits of the number be
x and y respectively.
And, the number be 10x + y = z
A → z+z2=25z
⟹ z = 24
B → (10y + x) – (10x + y) = 18
⟹y–x=2
10𝑥+𝑦
4
C → 𝑥+𝑦 = 1
⟹ y = 2x
Hence, either A alone or B and C together are sufficient to
answer the question.
1
St B — 𝑦 = 𝑥 ⟹ 𝑥 = 4𝑦
St. C —
4
36
𝑥−𝑦
=4 ⟹ 𝑥−𝑦 =9
So, any two of the three statements are sufficient to answer
the question.
S64. Ans.(d)
Sol. St A — Lengths = 4𝑥, 5𝑥
St B — ratio of speed = 1: 2
St C — speed of Ist train = 36 km/hr
From B and C
Speed of second train = 72 𝑘𝑚/ℎ𝑟
As we don’t know the directions of their motion so relative
speed can’t be determined
S65. Ans.(d)
√3
√3
B  MP-CP=Rs 480
C  MP = SP + 384
Hence, Any two of three statements are sufficient to answer
the question.
S60. Ans.(a)
Sol.
From A and B
A
:
B
:C
8100×10
: 10800×10
:
9
:
12
:
Profit share of B= Rs 1800
1800
Profit share of A=
× 9 = 𝑅𝑠 1350
12
A and B together only are sufficient to answer this question
S61. Ans.(b)
Sol. Sol. A  Profit percent = 25%
B  Let CP = 𝑥,
SP = 1.25𝑥
New CP = 𝑥 + 500
1.25𝑥−(𝑥+500)
100
Profit percentage =
× 100 = 9
𝑥+500
𝑥 = 4000
Profit = 1000 Rs.
C  C.P. = 𝑥
S.P. = 0.85(𝑥 + 1000)
0.85𝑥+850−𝑥
75
× 100 = 25 − 4
𝑥
𝑥 = 4000
Profit = (5000 − 4000) = 1000 Rs.
So A and either B or C are sufficient.
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S63. Ans.(e)
Sol. Let the speed of boat in still water and speed of stream be
x and y respectively.
45
St A — 𝑥 + 𝑦 = 3 ⟹ 𝑥 + 𝑦 = 15
Sol. St. C — 4 𝑎2 = 16√3, from here side of the equilateral
triangle and height can be calculated.
48
St. B — Side of triangle = 3×2 = 8
S59. Ans.(b)
Sol. Let the cost price be Rs 100x
Then Marked price=100𝑥 × 1.6 = 𝑅𝑠 160𝑥
A  SP = 112% of CP
4
7
And SP= 160𝑥 × 5 × 8 = 𝑅𝑠 112𝑥
52
S62. Ans.(e)
Sol. As we don’t know the time for which Rinku borrowed the
amount, so the rate of interest can’t be determined
ℎ= 2 𝑎
St. A — no conclusion
So using either B or C alone we can find the height.
S66. Ans.(b)
Sol. Quantity I
Let the number be 10x +y
Acc. to question
y=x+2
and
(10x +y) (x+ y) = 144
(10x + x+ 2) (x+ x + 2) =144
(11x+ 2) (x +1) = 72
11x² + 13x + 2= 72
11x² + 13x – 70 = 0
11x² + 35x – 22x – 70 = 0
On solving x = 2
Number is 24
Quantity II > Quantity I
S67. Ans.(b)
Sol. Quantity I
Let they meet after ‘n’ days
Applying Arithmetic progression
n
n
[2 × 15 + (n − 1)(−1)] + [20 + (n − 1)2] = 165
2
n
2
2
[30 − n + 1 + 20 + 2n − 2] = 165
n² + 49n – 330 = 0
n = –55, +6
so, they will meet in 6 days
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Quantity II
Let required no. of days
Quantity II:
Let first we arrange 4 men in 3! Ways, then 4 women can be
arranged in 4 places in 4P4 ways
= 3! × 4P4 = 144
= 144 ×8
= 1152
Quantity I= Quantity II
Days
Quantity II > Quantity I
S71. Ans.(a)
Sol. From I →
X + Y + Z = 16 × 3 + 12 = 60
Let, Y's 𝑎𝑔𝑒 𝑏𝑒 ′𝑎′ years
Then, X= a –d
Z = a+ d
⇒ 3a = 60, a = 20
Y’s present age = 20 years
(X + Z)’s present age = 60 – 20 = 40 years
From II →
Let x, y and z years be the respective age of X, Y and Z
𝑥+𝑧
=𝑦
2
x + z = 2y
z–y=4
x + z = 2(z – 4)
⇒ x + z = 2z – 8
⇒ x – z = -8
⇒z–x=8
Statement I alone is sufficient to answer the question but
statement II alone is not sufficient to answer the question.
S68. Ans.(a)
Sol. Quantity I:
Let present age of Randy = x
x−10
12
= 24 − 19
x – 10 = 5×12
x = 70 years
Quantity II:
Required average
=
=
14×
111
−2×42
4
12
777
−84
2
12
609
203
= 24 = 8
= 25.375 year
Quantity I > Quantity II
S69. Ans.(a)
Sol. Quantity I:
Let C.P of 100 gm = 100 Rs
So, he purchases 120 gm in 100 Rs
105
And sell 90 gm in = 100 × 100 RS
S72. Ans.(d)
Sol. According to concept if ratio between speed of two men
is a : b then their distinct meeting points are (b – a)
From I → Speed of Prabhat is 24 km/hr more than speed of
Rahul. By this we can’t conclude, will they meet at half of the
distance or not.
From II → Ratio between speed of Rahul to speed of Prabhat
→1:3
Distinct meeting points → (3 – 1) = 2
From statement (II) we can conclude, that they will meet at
half of the distance on a circular track
Hence, Statement II alone is sufficient to answer the question
but statement I alone is not sufficient to answer the question.
So, % profit
=
=
=
S.P.−C.P.
C.P.
105 100
−
90 120
100
120
21 5
−
18 6
5
6
× 100
× 100
× 100 =
21−15
18
5
6
× 100
36
= 90 × 100
= 40% profit
Quantity II:
50% → 12 Rs
So, 100→ 24 Rs
So, 80% → 19.2
There will be 0% profit if the book were sold for Rs.4.8 more
Quantity I > Quantity II
S70. Ans.(e)
Sol. Quantity I:
Let first we arrange all 4 men in 4! Ways then we arrange 4
women in 4P4 ways at 4 places either left of the man or right
of the man.
= 4! × 4P4 + 4! × 4P4 = 2 × 576
= 1152
53
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S73. Ans.(d)
Sol. Let there are ‘x’ red and ‘y’ yellow balls
Where, x + y = 7
𝑦𝐶
2
From I → 10𝐶2 = 15
⇒
𝑦×(𝑦−1)
90
2
=
2
15
y (y – 1) = 12
y=4
From II →
3×𝑥
1
10 𝐶 = 5
2
⇒
3×𝑥
45
=
1
5
x=3
y=4
Either statement I or statement II by itself is sufficient to
answer the question.
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S74. Ans.(b)
Sol. From I →
Let, number of pens and pencils sold be 2x and 3x respectively
while cost price of a pen and a pencil be 3y and 2y
respectively.
Total cost price =2x × 3y + 3x + 2y
= 12xy
137.5
Total selling price = 12𝑥𝑦 ×
= 16.5𝑥𝑦
100
From II →
Profit % on selling one pen and one pencil is 25% and 50%
respectively.
If overall profit % is 37.5%
⇒ Using allegation
No. of day required to complete the remaining work
800
=
50 + 3x
800
50 + 3x
= 10
x = 10
10 additional women are required to complete the remaining
work in 10 days.
Quantity II:
Let the additional number of men required be y.
There are 4 + y men and 10 women now.
Per day work of 4 + y men and 10 woman = (4 + y) × 5 + 10 ×
3 = 50 + 5y units
800
No. of day required to complete the remaining work = 50 + 5y
800
50 + 5y
≤8
y ≥ 10
At least 10 additional men are required to complete the
remaining work in either 8 or less than 8 days.
Quantity II ≥ Quantity I
Total cost price of pen = Total cost price of pencil
S75. Ans.(e)
Sol. Quantity I:
Let vessel A contains 3x litres milk and x litres water and
initial quantity of mixture in vessel A be 4x litres.
Half of the content of vessel A is first poured into vessel B, then
content of vessel B is poured into vessel C and finally contents
of vessel C is poured into vessel A.
So, vessel A finally contains contents of all the three vessels.
Final ratio of milk and water in vessel A:
Quanity of milk in all three vessels
Quanity of water in all three vessels
3x + 30
9
x + 20
9
∴
(8M+14W)×x×7
7
×360
12
=
(6M+10W)×15×6
5
×360
12
171
𝑥 = 13
2
= 13 13
2
∴ (Quantity I > Quantity II)
⟹ x = 20
Initial quantity of mixture in vessel A = 4x = 80 litres
Quantity I = Quantity II
S78. Ans.(b)
Sol.
S76. Ans.(d)
Sol. 20 men can complete the work in 12 days. So, 1 man can
complete the same work in 240 days.
Efficiency of 5 women = Efficiency of 3 men
5W = 3M
Ratio of efficiencies:
M
5
= 3
W
Let, a man does 5 units and a woman does 3 units of work per
day
& total units of work are 1200 units.
8 days’ work of 4 men and 10 women = 8 × (4 × 5 + 10 × 3) =
400 units
Remaining work = 1200 – 400 = 800 units
Quantity I:
Let the additional number of women required be x.
There are 4 men and 10 + x women now.
Per day work of 4 men and 10 + x woman
= 4 × 5 + (10 + x) × 3 = 50 + 3x units
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𝑥 = 10
∴ Quantity I → 10 + 5=15 days
Given 2M = 3W
Quantity II → 13 13
=4
= 4
54
S77. Ans.(a)
Sol.
A
B
C
Time 𝑥 + 5 𝑥 𝑥 − 4
1
1
1
∴ 𝑥+5 + 𝑥 = 𝑥−4
|
Let P is faster than Q
Then P covers 72 km distance in the same time as Q covers 48
km distance
Ratio of the speed = 72 : 48
=3:2
48
∴Speed of faster train i.e., P = 2 × 3 = 72 km/hr
Quantity 1→ Difference between P and Q = 72 – 48 = 24
km/hr.
Let speed of train = T km/hr
Let speed of car = C km/hr
120
480
∴
+
= 8 ………..(i)
𝑇
200
𝐶
400
1
+ 𝐶 = 8 3 ………(ii)
𝑇
On solving (i) and (ii)
T = 60 km/hr
∴ Quantity I < Quantity II
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S79. Ans.(a)
455 → 216 × 455 = 22750 Rs.
S82. Ans.(b)
Sol. Let no. of questions attempted correctly by him = x
No. of questions attempted wrongly by him = y
And no. of questions unattempted = z
Then,
x + y + z = 100 …(i)
4x – 2y – z = 320 …(ii)
From (i) & (ii),
6y + 5z = 80
80 –5𝑧
or, y =
Quantity I → 22750 Rs.
max. value of y =
1
Sol. 20% = 5
180 → 216
5×36 → 6×36
25×6 → 36×6 ⇒ 150 → 216] Installments
125 → 216
125 → 216
Total Principal = 455
∴ 216 → 10800
10800
6
80–5×4
6
60
= 6 = 10
1
And, 5% = 20
S83. Ans.(d)
Sol. Let, total work = 48 units
And efficiency A, B and C be a, b and C units respectively
Then,
a+b=4
b+c=3
and,
b=2
hence, c = 1 and a = 2
A and B have same efficiencies
Work done on day one= 2+2=4 units
Work done on day two= 1 unit
Work done in two days = 5 units
Work done in 18 days= 45 units
3
Remaining three units will be done by A and B in 4 days
20×441 → 21×441 8820 → 9261
400×21 → 441×21 ⇒ 8400 → 9261] Installments
8000 → 9261
8000 → 9261
Sum = 25220 Rs.
∴ 25220 → 25220
25220
9261 → 25220 × 9261 = 9261 Rs.
240
Quantity II = 100 × 9261
= 22226.4 Rs.
∴ Quantity I > Quantity II
S80. Ans.(b)
3
Hence, total no. of days taken=184 days.
6
Sol. Quantity I → 2 = 3
S84. Ans.(b)
Sol. Let the MP of a chair and a table be Rs.5x and Rs.8x
respectively.
And, the number of chairs and tables bought be 6y and 5y
respectively.
CP of a chair for Abhishek = (100 – 20) % of 5x = Rs.4x
CP of a table for Abhishek = (100 – 25) % of 8x = Rs.6x
Total CP for Abhishek = 4x × 6y + 6x × 5y = 24xy + 30xy = 54xy
SP of a chair for Abhishek = (100 – 25) % of (100 + 50) % of
4x = 4.5x
SP of a table for Abhishek = (100 – 20) % of (100 + 50) % of
6x = 7.2x
2
Number of chairs sold in bunch of four by Abhishek = 3 rd of
6y = 4y
1
So, number of table sold for free by Abhishek = th of 4y = y
4
Total SP for Abhishek = 4.5x × 6y + 7.2x × (5y – y) = 27xy +
28.8xy = 55.8xy
55.8xy − 54xy
1.8xy
1
Profit % =
× 100 = 54xy × 100 = 3 3 %
54xy
1
And, 3 𝜋𝑟 2 ℎ = 16𝜋
ℎ=3
∴ 𝑦 = √42 + 32 = √16 + 9
𝑦=5
∴ Quantity I < Quantity II
S81. Ans.(a)
Sol. Quantity of impurity in initial mixture
= 5% of 2kg = 0.1 kg
And let x kg of pure chili powder is mixed
ATQ
0.1
2+𝑥
4
= 100
𝑥 = 0.5 𝑘𝑔
0.1
Profit %=2+0.5 × 100 = 4%
S85. Ans.(c)
Sol. According to the question,
MP of a table = 300 + MP of a chair
⟹ 8x = 300 + 5x
⟹ x = 100
Total CP for Abhishek = 108000
⟹ 54xy = 108000
⟹ 54 × 100 × y = 108000
⟹ y = 20
Number of chairs purchased by Abhishek = 6y = 120
To earn 30%, let he marks-up by x%,
Then
(4 + 𝑥 +
4×𝑥
100
) % = 30%
104𝑥
4 + 100 = 30
104𝑥
100
= 26
or, x = 25%
55
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S86. Ans.(b)
Sol. Let, initially planned tons of ore to mine per day be ‘𝑥’ and
planned no. of days be ‘9𝑛’
9𝑛𝑥 = 1800 …………..(i)
ATQ,
3𝑛(𝑥 − 20) + (6𝑛 − 1)(𝑥 + 20) = 1800 …………(ii)
or, 3𝑛𝑥 − 60𝑛 + 6𝑛𝑥 + 120𝑛 − 𝑥 − 20 = 1800
or, 60𝑛 − 𝑥 − 20 = 0 ……………(iii)
200
putting𝑛 =
from (i) in (iii),
𝑥
𝑥 2 + 20 − 12000 = 0
⇒ (𝑥 + 120)(𝑥 − 100) = 0
⇒ 𝑥 = 100
S87. Ans.(c)
Sol. Tons of ore mined till one-third of planned no. of days = 6
× 100 = 600
Two days prior to the planned date means, the task has to be
completed in 16 days.
1
Tons of ore needed to be mined per day after rd period =
1800−600
16−6
S90. Ans.(b)
Sol. Let length of longer train and smaller train be 8x m and
7x m respectively
And that of speed of longer and smaller train be 3y m/s and
4y m/s respectively
ATQ—
(8x+7x)×7
(3y + 4y) =
90
15x = 90y
x = 6y …………… (i)
And
7x+(7x−100)
4y =
16
64y = 14x – 100………………… (ii)
From (i) and (ii)
y= 5
and x=30
speed of longer train= 15m/s
and length of longer train =8 × 30 = 240 𝑚
240+110
1
required time=
= 23 𝑠𝑒𝑐
15
3
1200
= 10 = 120
Extra tons of ore per day = 120 – 100 = 20
S88. Ans.(a)
Sol. Let the number of students in A, B and C be b, b and c
respectively. Let the total marks obtained by the students of
class A, B and C be x, y and z respectively. Total marks
obtained by the students of class A and B, B and C and A and C
= 2(Total marks obtained by the students of class A, B and C)
= 2(60(b + b + c))
= 120(b + b + c)
This also equals
52.5(b + b) + 70(b + c) + 60(b + c)
∴ 120(2𝑏 + 𝑐) = 235𝑏 + 130𝑐
b = 2c
Total number of students of A, B and C = b + b + c = 50
2(2c) + c = 50
⟹c = 10
S89. Ans.(e)
Sol. Number of students in all of the three sections- 20+20+10
=50
3
S91. Ans.(c)
Sol. From I,
Side of square = 24 cm
Radius of cone = Side of square – 3
r = 24 –3
r = 21 cm
From II,
66
7
Radius of circle = 22 × 2 = 10.5 𝑐𝑚
Height of cone = radius of cone + 7 .5cm
= 10.5 + 7.5 = 18 cm
From I & II
1
22
Volume of cone =3 × 7 × 21 × 21 × 18
= 8316 cm3
So, From I and II we can determine the area of cone.
S92. Ans.(d)
Sol. From I,
Let man invested Rs 100
137.5
So amount after three years = 100 × 100 = 𝑅𝑠. 137.5
Interest = 137.5 − 100 = 𝑅𝑠. 37.5
37.5 ×100
R = 100×3
R = 12.5%
From II,
Total interest = 13200 − 9600 = 3600 𝑅𝑠.
3600×100
R=
9600×3
R = 12.5%
So, either from I or II we can determine rate of interest.
S93. Ans.(d)
Sol. Let length of two trains be 9x meters and 8x meters
From I,
(90 + 72) ×
5
18
=
(9𝑥+8𝑥)9
68
153x = 3060
x = 20 meters
Required difference = 20 × 9 − 20 × 8 = 20 𝑚𝑒𝑡𝑒𝑟𝑠
56
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From II,
Let length of smaller train is L meters
5
𝐿
90 × 18 = 6.4
L = 160 meters
160
Length of larger train =
× 9 = 180 𝑚𝑒𝑡𝑒𝑟𝑠
8
Required difference = 20 meters
So, either I or II alone sufficient to give answer of question.
S94. Ans.(b)
Sol. Quantity I –
Total number of balls in bag = (5 + 6 + a + b) = (11 + a + b)
ATQ –
𝑎
1
=
(11+𝑎+𝑏)
6
6a – a - b = 11
5a − 𝑏 = 11---------- (i)
𝑏
2
Also,
=
11+𝑎+𝑏 9
9b = 22 + 2a + 2b
− 2𝑎 + 7𝑏 = 22 ----------------- (ii)
From (i) & (ii)
a=3
b=4
5×3
Required probability = 18×17 ( 2!)
= 20
240
𝑦 = 80
y = 3km/hr
x = 3 × 3 = 9 km/hr
240
Required time = (9+3) = 20 hours
6𝑥
32
= 3
64x = 3840
x = 60 km/hr
1200
Required time = 60 = 20 ℎ𝑜𝑢𝑟𝑠
Quantity I = Quantity II
20
20x = 56 + 7x + 7y
13x − 7𝑦 = 56 ------------------ (i)
𝑦
1
Also, 8+𝑥+𝑦 = 4
4y = 8 + x + y
𝑥 + 3𝑦 = 8 ---------- (ii)
From (i) & (ii) we get
y=5
x=7
8×7
Required probability = 2!× 20×19
28
= 95
Quantity I > Quantity II
S95. Ans.(e)
Sol. Quantity I –
22
Volume of cylindrical vessel = 7 × 17.5 × 17.5 × 18
= 17325 cm3
80
Volume of milk = 17325 ×
= 13860 cm3
100
30 × 7 × 3 × ℎ = 13860
462
h = 21
h = 22 cm
Quantity II –
Let length of rectangle and side of square be 12x & 11x
respectively
4× 11𝑥 − 2(12𝑥 + 18) = 4
44x – 24𝑥 = 4 + 36
x = 2 cm
Side of square = 2 × 11 = 22 𝑐𝑚
Quantity I = Quantity II
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4𝑦
1440+2400
= 51
Quantity II –
Total dice = 8 + x + y
𝑥
7
=
57
96+144
Quantity II –
Let speed of car be x km/hr
ATQ –
240 480×5 32
+ 6𝑥 = 3
𝑥
5
8+𝑥+𝑦
S96. Ans.(e)
Sol. Quantity I –
Let still water speed of boat x km/hr and speed of current y
km/hr
ATQ—
(x + y) = 2 (x – y)
x = 3y
ATQ—
96
72
+ (3y–y) = 20
(3y+y)
|
S97. Ans.(e)
Sol. From statement (A)
It can be observed that he obtained Rs. 900 in 3 years which
means Rs. 300 in 1 year and Rs. 300 is 10% of Rs. 3000.
Therefore,
A + 3000 < 5000
A < 2000
It can’t be solved further
From statement (B)
We can conclude that
A + 4000 > 5000
A > 1000
From statement (C)
A = either of 500, 1000, 1500……..
By combining all of the statement we will get
A = 1000 or 1500
Hence we can’t tell exact value of A.
S98. Ans.(e)
Sol. Let us assume selling price of item A and B is A and B
respectively, while cost price of item A and B is x and y
respectively. We have to calculate value of B.
From statement (A)
4A + B = 70
7x = 70

x = 10
and
4y + x = 70

y = 15.
But we can’t calculate value of B
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From statement (B)
We obtain 4 equations from this statement
B=y+3
13x = 10 A
B-A=5
y–x=5
here are 4 equations and 4 variables, by solving we can
get exact value of B as Rs. 18.
From statement (C)
Now profit % of both the items is 3 : 2. Let profit % be 3m%
and 2m% for item A and item B respectively.
From only C we cannot solve the question.
By using A and C:ATQ
4 × 10 ×
(100+3𝑚)
100
+ 15 ×
(100+2𝑚)
100
= 70
M=10
So value of B = 18
After combining statement (A) and (C) we can solve the
question.
Hence question can be solved by (B) alone or by using (A)
and (C).
S99. Ans.(c)
Sol. (A) a + b = –3
There are many cases possible
a = – 4, b = 1 ⇒ |a × b| < 10
a = –7 b = 4 ⇒ |a × b| > 10
(B)
a×b<0
Either a > 0 & b < 0
OR
b>0&a<0
(C)
a + 2.5b = 0
a
b
x
2
2
2k
3
2k
8k
= =
y
5k
say.
+2= 3
15k + 12 = 16k
k = 12
Hence age of Monu = 2k = 24 years.
Hence, we can answer from either two statements.
8×8
2 year CI on 8% = 8 + 8 + 100
Let a = – 5x & b = 2x
Minimum possible integers, which satisfies the eqn. are
When x = 1, a = –5 and b = 2
⇒ |a × b| = 10
or when x = 2
a = –10, b = 4
|a × b| > 10
Hence we can answer question from statement (C) alone.
S100. Ans.(d)
Sol. From (A)
Age of Sonu is 36
And their total age is 96 years.
We can’t tell more.
From (B)
Let age of Monu = 2x, age of Jonu is 3x.
Now their average age of all three could be
or
From (A) and (B)
We can tell age of Monu, as he is 33 ½% younger than Jonu,
and it is given in the question 2 of them have same age.
From (A) and (C)
Average age is 32 years.
Total age is 96 years.
Monu is youngest,
And age of Sonu and Jonu is y i.e. 36 years.
From (B) and (C)
Let age of Monu is x years.
Age of Sonu & Jonu is y years.
= 8%
Principle invested in UCO bank
–5
3
OR
x+y+y
y+2= 3
S101. Ans.(a)
Sol. Given, rate offered by IDBI : rate offered by UCO = 3 : 4
6
Rate offered by UCO bank = 3 × 4
=2
2x+3x+2x
From (C)
Let age of Monu is x
And age of Sonu and Jonu is y
ATQ,
x+y
x+y+y
+2= 3
2
= 16.64%
29160
Principle = 116.64 × 100
= 25000 Rs
10000×2×6
Amount obtained from IDBI =
+ 10000
100
= 1200 + 10000
= 11200
Required difference = 25000 – 11200
= 13800 Rs.
S102. Ans.(c)
Sol. Rate = 10%
According to question
Principle invested in SBI =
26250
100+4×10
× 100
= 18750 Rs.
Amounts obtained from Yes bank
3 years CI on 10% = 33.1%
133.1
= 20000 × 100 = 26620 Rs.
18750
Required% = 26620 × 100
2x+3x+3x
= 70.435%
3
We can’t tell further.
58
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S103. Ans.(b)
Sol. Principle invested in ICICI
20000
= 5 ×7
= 28000 Rs
Time = (3 – 1) = 2 year
15×15
2 year CI on 15% = 15 + 15 + 100
= 32.25%
32.25
Required interest = 28000 × 100
= 9030 Rs.
S107. Ans.(a)
Sol. Let Laddu’s prepared on Friday be 100x
Then Laddu’s sold on Friday be 80x
Also
Laddu’s prepared on Monday be 100y
Then Laddu’s sold on Monday be 90y
ATQ,
100𝑥+90𝑦
113
=
100𝑥+100𝑦 120
𝑥
S104. Ans.(d)
Sol. Let principle invested in UCO is X Rs. and principle
invested in ICICI is (x + 3000) Rs
According to question
15×15
2 year CI on 15% = 15 + 15 + 100
= 32.25%
8×8
2 year CI on 8% = 8 + 8 + 100
= 16.64%
132.25(𝑋+3000)
16.64𝑥
− 100 = 32870
100
115.61X = 3287000 – 396750
115.61x = 2890250
2890250
X= 115.61 = 25000 𝑅𝑠.
Principle invested in ICICI = 25000 + 3000
= 28000 Rs.
S105. Ans.(c)
6
Sol. Rate offered by SBI = 3 × 5
= 10%
4
Time = 2 × 1 = 2 𝑦𝑒𝑎𝑟
Principle invested in SBI
26250
= 100+(10×4) × 100
S108. Ans.(c)
Sol. Let Barfi prepared on Friday be = 100a
Laddu prepared on Friday = 100a + 80
Let Barfi prepared on Monday = 100b
ATQ,
80
[100𝑎 + 80] = 60b
100
100a + 80 = 75b
15b – 20a = 16 …(i)
Also,
100b 10
= ⇒ 7b = 10a or 14b = 20a
100a
112
= 10000 × 100
= 11200 Rs.
Required sum = 11200 + 18750
= 29950 Rs.
S106. Ans.(e)
Sol. Let he prepared 300 x kg Laddu on Saturday, then
Laddu’s sold on Saturday = 225x
Laddu’s prepared on Sunday = 500x
Laddu’s sold on Sunday = 400x
Barfi’s prepared on Saturday and Sunday each
500𝑥+300𝑥
=
= 400x
2
Barfi’s sold on Saturday = 280x
Barfi’s sold on Sunday = 320x
ATQ,
(400x + 225x) – (280x + 320x) = 100 kg
⇒ 25x = 100
x=4
therefore, his Laddu’s remained unsold on Sunday = 100x =
400 kg.
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And
90y × 20 – 80x × 20 = 11040
1800y – 1600x = 11040
90y – 80x = 552 …(ii)
On solving (i) and (ii) we will get
y = 16.8
100y = 1680 kg.
7
From (i)
b = 16
or
a = 11.2
Therefore, Barfi prepared on Friday = 1120 kg
= 18750 Rs.
Amount obtained from IDBI
100+(2×6)
= 10000 × 100
59
5
⇒ 𝑦 = 7 …(i)
|
S109. Ans.(a)
Sol. Let he prepared 500x kg of Laddu & 400x kg of Barfi on
Saturday.
Quantity of Barfi sold = 280x kg
120x
Profit from Barfi = 280x × 10 – 0.8 × 10
= 2800x – 1500x = 1300x
Quantity of Laddu sold = 375x
125x
Profit from Laddu = 375x × 10 – 0.8 × 10
= 3750x – 1562.5x
= 2187.5x
Required profit %
1300x+2187.5x
=
× 100
900x×200
3487.5x
= 1800x = 1.9375 ≈ 2%%
S110. Ans.(b)
Sol. Let he prepared 100a kg of Barfi on each day.
The laddu’s prepared = 75a, 125a and 105a.
Barfi sold on these 3 days = 70a + 80a + 60a = 210a
Laddu sold on these 3 days
225
=
a + 100a + 94.5a
4
= 56.25a + 100a + 94.5a = 250.75a
460.75
Required % =
× 100 = 76.15%≈ 76%
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S111. Ans.(a)
Sol. Let, one day work of a man, woman and youngster be
𝑚, 𝑤 and 𝑦 units,
ATQ, 2𝑚 = 3𝑤 = 4𝑦
One-day work of 14 men, 12 women and 12 youngsters
= 14𝑚 + 12𝑤 + 12𝑦
2
2
= 14𝑚 + 12 × 𝑚 + 12 × 𝑚
3
4
= 14𝑚 + 8𝑚 + 6𝑚
= 28𝑚
Total work = 24 × 28 𝑚 units
To finish it in 14 days,
24×28𝑚
= 48𝑚 units must be done daily.
14
Which means, 48𝑚 − 28𝑚 = 20𝑚 additional units are to be
done.
For this, 20 more men are required.
S112. Ans.(c)
1
Sol. To finish it in 19 = 19.2 days,
24×28𝑚
19.2
Solution (116-120)
Let, Number of farmers producing RICE = 5x
120
⇒ Number of famrers producing CORN = 5𝑥 ×
= 6𝑥
100
Let, Number of farmers producing only RICE and only CORN =
a
3
Farmers producing only WHEAT = 5𝑥 × 5 = 3𝑥
Let, Number of farmers producing only WHEAT and RICE = y
⇒ Number of farmers producing only WHEAT and CORN but
not RICE = 2y
But Number of farmers producing only RICE is half of number
of farmers producing RICE.
⇒ 𝑎 = 2.5𝑥
Number of farmers producing CORN is 48% of total farmers =
6𝑥
⇒ 100% = 12.5𝑥 = Total number of farmers
Number of farmers producing RICE and CORN
12
=
× 12.5𝑥 = 1.5𝑥
100
5
= 35 m units are to be done daily.
Which means, 35𝑚 − 28𝑚 = 7𝑚 additional units are to be
done daily.
2
1
7
Now, 1. 𝑤 + 1. 𝑦 = 3 𝑚 + 2 𝑚 = 6 𝑚
7𝑚
Daily work of a pair of women & youngster is 6 units
Hence, 6 such pairs are needed.
⇒ 1.5𝑥 + 2.5𝑥 + 𝑦 + 3𝑥 + 2𝑦 + 2.5𝑥 = 12.5𝑥
⇒𝑦=𝑥
Number of farmers producing only RICE and WHEAT but not
CORN = 24 = y
Put value of ‘y’ in above diagram.
By this we got,
S113. Ans.(c)
Sol. Profit on one articles = 30
Profit on 20 articles = 30 × 20 ⇒ 600
600 → 300
1 → 0.5
M.P. → 160 → 160 × 0.5 ⇒ Rs.80
S114. Ans.(e)
Sol. Let total quantity ⇒ 1000
He gives → 800 → for → 100
Cost price of → 800 → 80
30
Now initial discount =
× 100 = 18.75%
Reduced discount =
S.P. =
160×85
5
= 15%
= 136
100
(136–80)
Profit =
160
18.75×4
80
× 100 = 70%
S115. Ans.(b)
Sol. Let cost price of 20 articles → 20 × 100 = 2000
Actual profit on 20 articles → 20 × 30 = 600
S.P. → 2600
2600
S.P. of each undamaged article ⇒ 20–7 = 200
Initial discount =
30
160
× 100 = 18.75%
S116. Ans.(d)
Sol. Number of farmers producing at least two types of grain
= 24 + 16 + 20 + 48 = 108
Number of farmers producing at most one type of grain = 60 +
72 + 60 = 192
108
Required % =
× 100 = 56.25%
192
S117. Ans.(a)
Sol. Number of farmers producing WHEAT and CORN
= 16 + 48 = 64
125
Number of farmers producing Sugarcane = 64 × 100 = 80
200
So M.P. of each article should be = 81.25 × 100 ≈ 246
Approximately markup% = 146%
S118. Ans.(c)
160
1
Sol. Required % =
× 100 = 53 %
300
60
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S119. Ans.(b)
Sol. Number of farmers producing wither only RICE or only
WHEAT = 60 + 72 = 132
S120. Ans.(e)
Sol. Required ratio =
(24+20+48)
16
92
23
= 16 = 4
S121. Ans.(d)
Sol. Total markers sold by A = 12% × 15,000 = 1800
1800
X marker sold by A = 9 × 4 = 800
Y marker sold by A =
Z marker sold by A =
1800
9
1800
9
× 3 = 600
× 2 = 400
Let C.P. of one marker = ‘x’
140
60
S. P. of X marker = 100 × x × 100 = 0.84x
140
80
S. P. of Y marker = 100 × x × 100 = 1.12x
S. P. of Z marker =
140
100
×x×
90
100
= 1.26x
Total C.P. = [800 + 600 + 400]x = 1800x
Total S.P. = 800 × 0.84x + 600× 1.12x + 400×1.26x
= 672x + 672x + 504x
=1848x
1848x−1800x
48x
Total Profit Percentage =
× 100 = 1800x × 100 =
1800x
Let, S.P. of each marker = 2x, 3x, 4x
Total marker sold by C
18
=
× 15000 = 2700
100
X, Y and Z marker sold by C = 9 : 7 : 9
= 972; 756; 972
Total S.P =972 × 2x + 756 × 3x + 972 × 4x = 8100x
Total S.P. of Y marker
756×3x×48600
=
= Rs. 13608
8100 x
S124. Ans.(b)
15
Sol. Total markers sold by ‘B’= 100 × 15000 = 2250
X, Y and Z markers sold by B
=3:4:3
= 675; 900; 675
Satish Veer
X markers sold = 60% 40%
= 405; 270
Y markers sold = 40% 60%
= 360; 540
Let S.P. of each X and Y marker = x, y
ATQ
405x + 360y = 8370 …(i)
270x + 540y = 9180 …(ii)
By solving (i), and (ii)
x = 10, y = 12
2
23%
S122. Ans.(b)
21
Sol. Total markers sold by E = 100 × 15000 = 3150
X, Y and Z sold by E = 3 : 2 : 1
= 1575; 1050; 525
Let S.P. of each marker sold by E
= x, 1.5x, 3x
Total S.P. = x × 1575 + 1.5x × 1050 + 3x × 525
= 4725x
= 47250
⇒ x =10
S.P. of x, y, z = 10, 15, 30
20
Total marker sold by F =
× 15000 = 3000
9
18
6
14
3
25
4
20
X type of Marker sold by C = 25 × 100 × 15000 = 972
X type of Marker sold by F = 12 × 100 × 15000 = 1000
E sold maximum number of X type of markers
S126. Ans.(a)
Sol. Total work = 10080 units (LCM of days taken by all)
10080
Efficiency of A =
= 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
70
10080
Efficiency of C =
Efficiency of D =
90
10080
32
= 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
= 315 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
7
New efficiency of A = 144 × 8 = 126 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
2
New efficiency of D = 315 × = 210 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
10080
3
Required time = (126+112+210)
= 22.5 ℎ𝑜𝑢𝑟𝑠
S123. Ans.(c)
Sol. Let, total C.P. = x
ATQ
10
8
x × − [x × ] = 9000
S127. Ans.(c)
Sol. Total work = 10080 units (LCM of days taken by all)
10080
Efficiency of A =
= 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
9
Efficiency of B=
2
x = 9000
9
Efficiency of E =
x = 40,500
Total S.P. of marks if C wants to earn 20% profit
120
= 40500 ×
= 48600
Efficiency of F =
70
10080
60
10080
48
10080
36
= 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
= 210 units/hour
= 280 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
3
New efficiency of F= 280 × 4 = 210 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
100
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X type of Marker sold by E = 6 × 100 × 15000 = 1575
X, Y and Z sold by F = 4 : 5 : 3
= 1000; 1250; 750
Total S.P. of markers sold by F
= 10 × 1000 + 15 × 1250 + 30 × 750
= 10,000 + 18,750 + 22,500
= Rs.51250
61
3
X type of Marker sold by B = 10 × 100 × 15000 = 675
X type of Marker sold by D = 15 × 100 × 15000 = 840
100
9
S125. Ans.(e)
4
12
Sol. X type of Marker sold by A = 9 × 100 × 1500 = 800
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ATQ –
(210+210 )(𝑦)
168(𝑦+1)
Total work in 2 hours = 802 – 427 = 375 units
10080×2
Total required time =
[ 375 is the total work done in 2
2
=1
1344
= 25
84
144
S130. Ans.(d)
10080
Sol. Efficiency of E = 48 = 210 units/hour
Efficiency of B =
2
Efficiency of C =
Efficiency of C =
Efficiency of D =
Efficiency of E =
Efficiency of F =
60
10080
90
10080
32
10080
48
10080
36
1680
= 210 units/hour
= 15 days
= 280 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
S131. Ans.(b)
Sol. Relative speed of train A and Man =
750
5
5
=
× + 10 ×
3
75
100
= 126 𝑢𝑛𝑖𝑡𝑠/hour
60
n = 4.5 ℎ𝑜𝑢𝑟𝑠
Total time = n + (n + 23.5)
= (4.5 + 4.5 + 23.5 )
= 32.5ℎ𝑜𝑢𝑟𝑠
S129. Ans.(d)
Sol. Total work = 10080 units (LCM of days taken by all)
10080
Efficiency of A =
= 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
Efficiency of D =
Efficiency of E =
Efficiency of F =
60
10080
90
10080
32
10080
48
10080
36
= 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
= 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
18
50
18
= 12.5 + 18
275
= 18 m/s
9
275
Length of train A = 150 m
Distance in 24 sec = 24 × 12.5
= 300 m
Length of platform = 300 – 150
= 150 m
S132. Ans.(c)
Sol
Let length of train C and train F be 3L and 5L respectively
Relative speed of train C to train F
2000
18
1000
18
= 3×60 × 5 + 60 × 5
= 40 + 60 = 100 km/hr
5
500
= 100 × 18 = 18 m/s
According to question
500
= 210 units/hour
L = 50 m
Length of train C and F
C = 3 × 50 = 150 m
F = 5 × 50 = 250 m
18
= 280 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
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Distance in 911 sec = 18 × 11 = 150𝑚
= 315 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
First hour total work of A, B, E and F
= (144 + 168 + 210 + 280)
= 802 units
In Second hour total task destroy by C & D
= −(315 + 112)
= − (427)
62
= 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
Required days = 112
Let C work for n hours and G & H work for (n + 23.5) hours
ATQ –
𝑛 × 112 + (216 + 126)(𝑛 + 23.5) = 10080
112n + 342(n+23.5) = 10080
112n + 342n + 8037 = 10080
112n + 342n = 10080 −8037
454n = 2043
2043
n = 454
Efficiency of C =
90
= 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
= 315 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
H in one hour = 168 ×
Efficiency of B =
60
10080
= 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
= 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
G in one hour = 144 × 2 = 216 units/hour
70
10080
10080
Work done by E & B:
E = 210 x 12 = 2520
B = 168 x 35 = 5880
Remaining work = 10080 – (2520 + 5880)
= 10080 - 8400
= 1680 units
S128. Ans.(a)
Sol. Total work = 10080 units (LCM of days taken by all)
10080
Efficiency of A = 70 = 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟
10080
19
= 53 25 hours
y = 4 hour
Total work = 420y + 168(y+1)
= 420 x 4 + 168(4+1)
= 420 × 4 + 168 × 5
= 2520 units
2520
1
A will complete =
= 17 ℎ𝑜𝑢𝑟𝑠
Efficiency of B =
375
hours ]
672 ×2
= 25
420y = 336y + 336
420y – 336y = 336
84y = 336
336
y=
𝑇=
=
|
3𝐿+5𝐿
= 14.4
150+(150+50)
2000
3×60
350×3×60
2000
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= 2 𝑠𝑒𝑐
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S133. Ans.(a)
Sol. According to question
180+180
18
Speed of train B = 21.6 × 5
S136. Ans.(b)
Sol. Let number of lectures taken by professor on Monday and
Tuesday each of second week be ‘x’.
ATQ,
3
1
5
2
[ × 8 × 2 + × 𝑥 × 2 + × 6 × 2 + × 9 × 2] × 4000
=
4
2
3
3
= 60 km/hr
Speed of train E in km/hr
300+300
18
𝐸 𝑠𝑝𝑒𝑒𝑑 = 30 × 5
600
18
= 30 × 5 = 72 km/hr
Relative speed of train E and B when running in same
direction
5
12×5
= 72 − 60 × =
18
10
18
= 3 m/s
Time taken by faster train to cross slower train =
=
480×3
10
(180+300)×3
10
= 144𝑠𝑒𝑐
2,00,000
⇒ 12 + x + 20 + 12 = 50
⇒x=6
S137. Ans.(c)
40
Sol. Required amount = [9 × 2 × 4000 + 8 × 2 × 6000 +
60
10 × 2 × 9000]
2
= [72,000 + 96,000 + 1,80,000]
3
2
= [3,48,000] = 2,32,000 = Rs. 2.32 lakh
3
S138. Ans.(e)
Sol. Let no. of lectures taken by professor on third week be ‘x’
ATQ,
45
1,21,500 = 6000 × [7 + 5 + 𝑥 + 8]
S134. Ans.(a)
Sol. Given,
Speed (m/s) : Taken time = 8x : 5x
120+240
8𝑥 = 5𝑥
60
⇒ 20 + x = 27
⇒x=7
7–5
Required % = 5 × 100
2
40𝑥 = 360
360
𝑥 = √ 40 = 3
2
= 5 × 100 = 40%
Speed of train D = 8 × 3 = 24 m/s
Taken time = 5 × 3 = 15 sec
120+600
S139. Ans.(b)
5×6000×45
Sol. Required ratio = 10×9000×40
24
Required ratio = (200+600)×60
=5∶8
1000
S135. Ans.(b)
Sol. Given,
Speed of smaller train D = 54 km/hr
Speed of train B = V km/hr
Relative speed = (V – 54) km/hr
To cross the man, who sits in smaller train D, train B have to
cross its own length with relative speed
5
180
= (𝑉 − 54) × 18 = 24
V = 81 km/hr
5
Required speed = 81 × 18
=3:8
S140. Ans.(c)
Sol. Let no. of lectures given on Monday of second week be ‘x’
[𝑥×4000×2+5×6000×2+6×9000×2]60
112.5
= [8×4000×2+7×6000×2+6×9000×2]45
100
9
4 [8000𝑥+1,68000]
⇒8=3
27
=
32
[2,56,000]
8000𝑥+1,68,000
2,56,000
⇒ 8000x = 2,16,000 – 1,68,000
⇒ 8000x = 48000
⇒x=6
45
= 2 m/s
S141. Ans.(d)
Sol. Let cost price of article X on Monday and Thursday be
400x and 500x respectively.
Selling price of article X on Thursday
175
60
= 500𝑥 × 100 × 100 = 525𝑥
ATQ,
525𝑥 − 500𝑥 = 75
75
⇒ 𝑥 = = 3 Rs.
25
Mark price of article Y on Monday
180
5
= 400 × 3 × 100 × 4 = 2700 Rs.
Selling price of article Y = 2700 ×
65
100
= 1755 Rs.
Required profit = 1755 − 400 × 3
= Rs 555
63
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S142. Ans.(b)
Sol. Let cost price of article X and Y on Friday = 100x
150
90
Selling price of article X = 100𝑥 ×
×
100
Profit on selling article X on Friday
150
90
= 500𝑥 × 100 × 100 − 500𝑥= 175x
100
= 135x
Selling price of article Y = 135x + 520
ATQ,
135x – 100x + 135x + 520 – 100x = 1430
⇒ 70x = 910
⇒ x = 13
150
7
Mark price of article Y on Friday = 100𝑥 ×
× = 210𝑥
100
= 210 × 13 = 2730
Required % =
2730−[135𝑥+520]
2730
455
5
× 100
2
= 2730 × 100 = 16 3 %
S143. Ans.(e)
Sol. Let cost price of article X and article Y on Wednesday be
300x and 400x
126
Mark price of article Y on Wednesday = 400𝑥 ×
+ 1008
100
= 504𝑥 + 1008
165
14
Mark price of article Y also equals to = 300𝑥 × 100 × 11 =
630𝑥
⇒ 630x = 504x + 1008
1008
⇒ 𝑥 = 126 = 8
165
S150. Ans.(c)
Sol. Maximum number of bikes produced = 750 , on Thursday.
S144. Ans.(c)
Sol. Let cost price of article X = 100x
Cost price of article Y = 80x
150
S151. Ans.(b)
Sol. ATQ,
4
Mark price of article Y on Tuesday = 100𝑥 × 100 × 3 = 200𝑥
Selling price of article Y on Tuesday = 200𝑥 ×
120𝑥−80𝑥
80𝑥
40
60
100
= 120𝑥
× 100 = 80 × 100 = 50%
S145. Ans.(d)
Sol. Let cost price of article X on Monday, Tuesday,
Wednesday, Thursday and Friday be 100x, 200x, 300x, 400x
and 500x respectively.
Profit on selling article X on Monday
180
80
= 100𝑥 × 100 × 100 − 100𝑥 = 44x
Profit on selling article X on Tuesday
150
80
= 200𝑥 × 100 × 100 − 200𝑥= 40x
70
100
R
2
= 10,000 [1 + 100]
⇒ 320R = 10,000 + R² + 200R – 8000
⇒ R² – 120R + 2000 = 0
⇒ R² – 100R – 20R + 2000 = 0
⇒ R (R – 100) – 20 (R – 100) = 0
⇒ (R – 20) (R – 100) = 0
⇒ R = 20%, 100%
S152. Ans.(e)
Sol. Let he buy ‘x’ eggs out of which ‘y’ are rotten.
ATQ,
Total C.P. = 10x
108
Total S.P. = 100 × 10𝑥 = 10.8x
150
1080x = 800y + 1500x – 1500y
700y = 420x
𝑥
700 5
⇒ 𝑦 = 420 = 3
Profit on selling article X on Thursday
175
60
= 400𝑥 ×
×
− 400𝑥 = 20
100
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8000×4×R
10.8x = 10y × 100 + 10(x– y) × 100
= 300𝑥 × 100 × 100 − 300𝑥= 46.5x
64
8000 +
80
Profit on selling article X on Wednesday
100
S147. Ans.(a)
Sol. Total number of bikes produced by Bajaj from Monday to
Friday = 900
S149. Ans.(c)
Sol. No. of bikes produced on Tuesday and Thursday is same
i.e. 150
= Rs 2772
Selling price of article Y = 504x
= 504 × 8 = 4032
Required difference = 4032 – 2772 = 1260
165
520
70
= 300 × 8 × 100 × 100
Profit % =
Frida
y
180
140
200
S148. Ans.(e)
1150
Sol. Required average = 5 = 230
70
Selling price of article X = 300𝑥 × 100 × 100
165
Total profit = 44𝑥 + 40𝑥 + 46.5𝑥 + 20𝑥 + 175𝑥
= 325.5x
ATQ,
44x = 176
⇒x=4
Total profit = 325.5x = 325.5 × 4 = Rs 1302
Sol.
(146-150)
Mond Tuesd
Wednesd Thursd
ay
ay
ay
ay
Hero 180
150
250
150
Bajaj 160
220
200
180
Hond 200
200
300
250
a
540
570
750
580
S146 Ans.(b)
540
Sol. 750 = 18 ∶ 25
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Let total eggs = 5a
Rotten eggs = 3a
2a 2
Required probability = =
Alternative,
Use Allegation method
5a
S155. Ans.(b)
Sol. Let the total volume of tank is 100m.
Let efficiency of pipe B is 6x/hour
And that of pipe A is 8x/hour
Water flowing in the tank when all three pipes were opened
in an hour= 6x + 8x – 40
= 14x – 40 ltr. (This is 25% of volume of tank+ 10 liters)
13
13
Water filled in tank in initial
hours is 14𝑥 × =
5
7
7
26 𝑥 𝑙𝑖𝑡𝑒𝑟𝑠
Now this 14x-40 + 26x = 40x – 40 is total volume of tank.
25
Now 25 % of this total volume is 100 × (40x – 40)= 10x – 10
2
Required Probability = 5
S153. Ans.(a)
Sol. Area of hexagon = Area of 6 equilateral triangles = Area of
3 parallelograms = Area of 2 equilateral triangle + Area of 2
parallelograms
6
⇒ Area of parallelogram = 3 × 4√3 = 8√3
Let, side of equilateral ∆ = a
√3 2
a = 4√3
4
⇒a=4
Area of parallelogram = a × h
Where a = side of hexagon on equilateral triangle = 4 cm
and
h = height of parallelogram
So, 8√3 = 4 × h
⇒ h = 2√3 cm
S154. Ans.(e)
Sol. Total 100% work in completed by all four.
Let A, B, C and D worked (a – 2d), (a – d), (a)% and (a + d) %
work.
a – 2d + a – d + a + a + d = 100%
4a – 2d = 100%
But a = 30%
⇒ d = 10%
A, B, C and D completed 10%, 20%, 30% and 40% work.
C completes 30% work in 9 days.
C completes 100% work in 30 days.
B completes 20% work in 3 days.
B completes 100% work in 15 days.
A completes 10% work in 2 days.
A completes 100% work in 20 days.
Let D alone can complete 100% work in ‘d’ days.
4
4
4
4
+ 15 + 20 + d = 1
30
1
1
1
d
1
4
1
30
⇒ = –[
+
1
15
+
1
20
S156. Ans.(b)
Sol. Let A fills 8x/hour water.
While B fills 6x/hour water
When all pipes were opened together
They will fill (8x + 6x – 40) liters/hour.
14x – 40 liters is 10 liters less than 25% of tank’s efficiency.
It means 25% efficiency of the tank will be (14x- 30) liters.
100
(14𝑥 − 30) = 56x – 120.
Total efficiency of tank will be
25
In 4 hours, pipe A will fill 8𝑥 × 4 = 32𝑥 𝑙𝑖𝑡𝑒𝑟𝑠.
ATQ,
32x = 56x – 120
X = 5.
Hence volume of tank is 32×5= 160 liters.
S157. Ans.(d)
Sol.
80
speed of boat in still water = (10 + 10 × 100) km⁄hr
= 18 km⁄hr
ATQ—
280
280
560
+ (18+10)+s + (18–10)+s = 45
(18+10)
280
28+s
8
28+s
560
+ 8+s = 35
16
+ 8+s = 1
64 + 8s + 448 + 16s = 224 + 28s + 8s + s2
s² + 12x – 288 = 0
s = 12 km/hr
S158. Ans.(c)
Sol. Let investment of A, B, C and D is a, b, c and d respectively.
A
B
C
𝐷
]
Now in firt year → a × 12 : b × 12 : c × 12
⇒ d = 10
⇒ d = 10
D can complete 100% work in 10 days.
⇒ D can complete 40% work in 4 days.
65
ATQ,
10x – 10 + 10 = 14x – 40
4x = 40
X = 10 liters
Hence, volume of tank is 4x – 40 = 400 – 40 = 360 litres.
360
Hence, C will fill tank in 40 = 9 hours.
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In 2nd year
→ 2a × 12 :
4b
3
× 12 :
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5
6c
In 3rd year
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6c
5
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× 12
× 12 : 𝑑 × 12
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A:B:C:D
4
6
⇒ (a × 12 + 2a × 12) : (b × 12 + b × 12) : c × 12 + 2 c × 12 : d
S161. Ans.(b)
Sol. Difference between second year’s interests
× 12
= 12000 (1 + 7 2 %) (6 4 %) − 12000 (5 5 %)
3
7𝑏
5
17
1
43
3a : 3 : 5 𝑐 : d = 12 : 14 : 17 : 8
⇒a:b:c:d=4:6:5:8
Difference between B and C initial investment = 1150
Total Investment of A and D together
1150
=
× 12 = 13800
1
S159. Ans.(a)
Sol.
1
4
1
29
= 12000 (40) (16) − 12000 ( 5 %)
= 806.25 – 696
= Rs.110.25
S162. Ans.(c)
Sol. Loan of Rs.20480 is settled by paying Rs.27778 after five
years. Simple interest is applicable for first three years while
compound interest is applicable for next two years.
3
1
3
4
4
4
27778 = 20480 (1 + (8 % + 5 % + 4 %)) (1 +
1
x
7 2 %) (1 + 100)
3
1
x
4
2
100
27778 = 20480 (1 + (18 %)) (1 + 7 %) (1 +
19
43
)
x
27778 = 20480 (16) (40) (1 + 100)
1
For second work
11
Efficiency of A = 9 × 9 = 11 units/day
x = 64%
S163. Ans.(d)
Sol.
Let the amounts borrowed under plan B and C be 19x and 13x
respectively.
∴ Ratio of interests
9
Efficiency of B = 11 × 11 = 9 units/day
Work done by A = 11 × x = 11x units.
Work done by B = 9 × (x + 4) = (9x + 36) units
ATQ,
(9𝑥+36)–11𝑥
=
20𝑥+36
1
1
⇒ 828 – 46x = 20x + 36
⇒ 66x = 792
⇒ x = 12
Required time = (
20×12+36
20
1
3
= 16𝑥 × ((1 + 7 2 %) (1 + 6 4 %) − 1) : 13𝑥 × (8 4 % +
23
1
5 4 %)
43
17
14
= 16𝑥 × ((40) (16) − 1) : 13𝑥 × (100)
4
) = 13 days.
5
91
14
= 16𝑥 × 640 : 13𝑥 × 100
S160. Ans.(d)
Sol. Let original marked price be Rs 100x.
225
Then, New marked price of that article =100𝑥 × 100 =
=5:4
𝑅𝑠 225𝑥
Selling price of article
3
4
8
= 225𝑥 × × ×
For old plan C = 8 4 % + 5 4 % + 4 4 % = 18 4 %
4
5
3
9
85
8
8
S165. Ans.(a)
Sol. Let the amount borrowed under plan A be Rs.x
Effective rate of interests for three years:
2
2
2
For plan A = 8 % + 6 % + 3 % = 19%
5
− 96𝑥) = 255
3
4
3
3
5
5
5
Total interest = 19% of x + 18% of (30000 – x)
⟹ 5540 = 18% of 30000 + 1% of x
⟹ x = Rs.14000
x = 24
cost price = 96 × 24 = Rs 2304
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3
3
For plan E = 7 % + 5 % + 4 % = 18%
x = 255
66
× 100
3
8
875𝑥
3
18
4
2
875𝑥
24𝑥 − (
3
4
(20−18 )
=6 %
= 225𝑥 × 8 × 3 × 6
ATQ
3
% Increase in interest = % Increase in effective interest rate
for three years
= Rs 96x
Profit in first case=120𝑥 − 96𝑥 = 24𝑥
2nd selling price
= 𝑅𝑠
3
2
=
2
1
For new plan C = 3 × 6 3 % = 20%
= Rs 120x
C.P. of article
100
= 120x × 125
7
S164. Ans.(c)
Sol. Effective rate of interests for three years:
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S166. Ans.(c)
Sol. Candidate of party XYZ obtained
25
=
× 300000
100
= 75000 votes
Therefore, Total votes polled in constituency B
100
= 25 × 70000
S169. Ans.(b)
Sol. Votes scored by winner candidate in constituency E
=
S167. Ans.(a)
Sol. Votes scored by party XYZ from C
37.5
=
× 300000 = 112500
100
As party obtained 50% votes, therefore
Total votes polled in constituency E
= 2 × 112500 – 45000
= 225000 – 45000
= 180000
Votes scored by winner candidates from E
65
= 100 × 180000
= 117000
Votes scored by party from E
12.5
= 100 × 300000
= 37500
Required answer = 117000 – 37500 = 79500
=
12.5
100
× 300000
= 37500
They lost by 117000 – 37500 = 79500 votes
Hence winner candidate from constituency A got
= 30000 + 79500
= 109500
Hence votes polled in A
100
= 30 × 109500
≅ 365000
S170. Ans.(d)
Sol. (i)
If Party XYZ wins from C, total votes polled in constituency is
37.5
= 2 × 100 × 30000
= 225000
Candidate of party XYZ got 75000 votes from constituency B.
If these constituencies have equal votes, winner candidate
gets
25
= 100 × 225000
S168. Ans.(a)
Sol. Votes secured by party XYZ from A
10
= 100 × 300000 = 30000
Votes secured by party XYZ from D
15
= 100 × 30000 = 45000
If party wins from A, total votes from A
100
= 30 × 3000 = 100000
votes from D
winner candidate got = 45000 + 15000 = 60000 votes
100
Total votes polled in D = 40 × 60000 = 150000 votes
Difference = 150000 – 100000 = 50000
If party wins from D, total votes from D
100
= 45000 × 40
= 112500
& winner candidate from A got
= 30000 + 15000
= 45000 votes
Total votes polled in constituency A
10
= 30 × 45000
= 150000
Difference = 37500.
But maximum difference is of 50000, hence option a is the
answer.
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× 180000
= 117000
Votes scored by party XYZ
= 30000
Runner up candidate got = 75000 – 7000 = 68000
Now for maximum number of candidates, all of the other
should get min. possible votes
= Remaining votes = 300000 – [75000 + 68000] = 157000
Min. votes possible 12000
157000
Maximum candidates = 12000 ≈13
Hence total number of candidates is 13 + 2 = 15
67
65
100
|
= 57750
Hence it is not possible.
(ii) Party got, 37500 votes from E, which is 35% of total polled
votes (If we assume only 2 candidates). Hence winner must
have more than twice votes.
It is also not possible.
(iii) It is possible, and we can’t check it whether party lost or
won.
(iv) it is also possible, as we don’t have information on
number of votes.
S171. Ans.(a)
Sol. Total surface area of the toy = C.S.A of cone + C.S.A of
Hemisphere
Let, slant height of cone
πrℓ + 2πr² = 858 cm²
πr(ℓ + 2r) = 858 cm²
ℓ =25 cm
height of cone = 24cm
volume of the toy
1
2
= πr 2 h + πr 3
3
1
= 3 πr
3
2 (h
+ 2r)
2
= 1950 cm3
3
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S172. Ans.(e)
Sol. Height of cylinder = 12 cm
11
Height of toy is 4 𝑡ℎ 𝑜𝑓 the height of cylinder = 33 cm
Edges of cube = 33 – 12 = 21 cm
C.S.A of cylinder = 2πrh = 66 × h
r = 10.5 cm
Total surface area of toy = (6a² – πr²)+ 2πrh + πr²
(–πr², area subtracted due to aligment)
22
= 6 × 21 × 21 + 2 × × 12 × 10.5= 3438 𝑐𝑚2
7
S173. Ans.(c)
21
Sol. Sphere radius =
2
21
3
litre taxi P travel 1km
In 10 litre taxi P travel ‘x’ km
Price of petrol is Rs. 60/litre
60𝑥
So, Amount paid for petrol = 10 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑀𝑖𝑙𝑒𝑎𝑔𝑒
× 60
10
3
4
r+ h ∶ h ∶ 3r
13 : 6 : 7
12
Distance travelled by taxi ‘Q’ = 350 km
For taxi ‘R’
r
× 60 + r × 5 = 2592 (r= distance travelled by R)
15
S174. Ans.(d)
1
Sol. Volume of cone = 3 πr 2 h
1
1
10
𝑥
⇒ x = 240 km
Distance travelled by taxi ‘P’ = 240 km
Similarly
For taxi ‘Q’
q
× 60 + q × 5 = 3500 (q= distance travelled by Q)
height of cylinder = 12
required ratio
4
4
= πr 3 + πr 2 h ∶ πr 2 h ∶ πr 3
3
In
Total amount paid = Amount for petrol + 5 × Distance
travelled
x
∴ 𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑝𝑎𝑖𝑑 𝑡𝑜 𝑡𝑎𝑥𝑖 ‘P’ = × 60 + x × 5 = 2640
So, cylinder radius= 2
4
Required ratio = 2 : 7
Solution (176-180)
Let taxi P travelled x km.
So,
In 1 litre taxi P travel 10 km
Distance travelled by taxi ‘R’ = 288 km
For taxi ‘S’
s
× 60 + s × 5 = 4500 (s= distance travelled by S)
18
22
= 3 × 7 × 7 × 7 × 24 = 1232 cm2
Volume of cuboid
= 24 × 10 × 25% of 24
= 1440 cm²
Difference = 1440 – 1232 = 208 cm²
208
Required % = 1232 × 100 = 16.88% ≈ 17%
Distance travelled by taxi ‘S’ = 540 km
For taxi ‘T’
𝑡
14
× 60 + t × 5 = 2925 (t= distance travelled by T)
Distance travelled by taxi ‘T’ = 315 km
S175. Ans.(c)
Sol.
S176. Ans.(c)
Sol. Distance travelled by taxi ‘Q’ = 350 km
If millage of Q decreases by 2
So, amount paid to Q
=
350
12–2
× 60 + 350 × 5
= 3850
Distance travelled by taxi ‘S’ = 540 km
If millage of S decreases by 2
So, amount paid to S
540
= (18–2) × 60 + 540 × 5 = 4725
Total surface area of cone = πr × slant height + πr²
= πr (ℓ + r)
= πr (√8r 2 + r 2 + r)
= πr (4r)
= 4πr²
Total surface area of remaining part of cylinder
= πr² + 2πrh + πrℓ
= πr (r + 2h + ℓ)
= πr (r + 10r + 3r)
= 14 πr²
68
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Increase in amount = (3850 + 4725) – (3500 + 4500) = 575
S177. Ans.(d)
Sol. Distance travelled by Q = 350 km
350
Average speed of Q = 25 = 14 km/hr
Distance travelled by taxi ‘P’ = 240 km
Average speed of P =
Required % =
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15–14
15
240
16
= 15 km/hr
20
2
× 100 = 3 % = 6 3 %
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S178. Ans.(a)
Sol. Distance travelled by S = 540 km
540
Average speed of S = 30 = 18 km/hr
Let two speeds with which taxi S travel is 3x km/hr and 2x
km/hr
So,
2×3x×2x
3x+2x
12x
5
= 18
= 18
S182. Ans.(d)
Sol. Let present age of Simmi and Rimmi be 4x yr and 3x yr
respectively
And let present age of Rina be y yr.
ATQ
4𝑥+6
3
=1
𝑦
4𝑥 + 6 = 3𝑦……….. (i)
And
3𝑥
2
=
𝑦
18×5
x = 12 = 7.5
270
Time for higher speed =3×7.5 = 12 hour
Alternate,
Speed is inversely proportional to time if distance is same
So,
Higher speed : Slower speed = 3 ∶ 2
Time in Higher Speed : Time in lower speed = 2 ∶ 3
Total time = 30 hours
𝐼𝑓 5 → 30
30
⇒ 2 → 5 × 2 = 12 ℎ𝑜𝑢𝑟𝑠
1
⇒3x=2y …………. (ii)
From (i) and (ii)
y= 18 yr
present age of Simmi=48 yr
and that of Rimmi= 36yr
18+48+36
required average = 3
= 34 𝑦𝑟
S183. Ans.(d)
Sol. Let the value of x and y be 9a and 7a respectively
720×100
CP = (100+9a)
Now,
720×100
SP = (100+9a) & CP = Rs. 720
ATQ,
S179. Ans.(d)
Sol. Distance travelled by taxi ‘R’ = 288 km
Distance travelled by taxi ‘T’ = 315 km
720×100×100
(100+9a)×(100–7a)
= 720
⇒ 10000 = 10000 + 200a – 63a²
⇒ 63a² – 200a = 0
200
⇒ a = 0 or
a = 63
288
Average speed of R = 18 = 16 km/hr
315
Average speed of T = 15 = 21 km/hr
Required SP =
720×(100+7×
100
Ratio = 16 : 21
𝑥×100
Breadth of second field = (100+4𝑎) cm
ATQ,
x×100
(100–a)
240+350+288+540+315
5
= 346.6 km
S181. Ans.(e)
Sol. Let initial quantity of milk and water be 9x lit and 2x lit
respectively.
ATQ
9𝑥−44×
2𝑥−44×
9
11
2
+12
11
3
=1
3
New quantity of milk=9 × 16 − 36 + 64 × 8 = 132𝑙𝑖𝑡
5
And that of water=2 × 16 − 8 + 12 + 64 × 8 = 76 𝑙𝑖𝑡
132
69
y×100
× (100+4a) = xy
⇒ 10000 = (100 – a) (100 + 4a)
⇒ 10000 = 10000 + 400a – 100a – 4a²
⇒ 4a² – 300a = 0
⇒ a = 0 or 75.
S185. Ans.(b)
Sol. Let the amount of Aman and Vikash be Rs. 3x and 5x
respectively.
Total amounts = Rs. 8x
Total amounts after 2 years
20
⇒ 𝑥 = 16𝑙𝑖𝑡
Required ratio=
76
= 880
S184. Ans.(b)
Sol. Let length and breadth of first rectangular field is x cm
and y cm respectively
Area of first rectangular field = xy cm²
𝑥×100
Length of second field = (100–𝑎) cm
S180. Ans.(e)
Sol. Distance travelled
P → 240 km
Q → 350 km
R → 288 km
S → 540 km
T → 315 km
Required Average =
200
)
63
= 33: 19
2
20
= 3x × (1– 100) + 5x × (1 + 100)
4
4
6
6
5
5
5
5
2
= (3x × × ) + (5x × × )
= 1.92𝑥 + 7.2x = 9.12x
Change = Rs. 9.12𝑥– 8x = Rs.1.12𝑥
1.12𝑥×100
% change =
= 14%
8𝑥
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S186. Ans.(c)
Sol. On Tuesday
Gaurav =
25×100
Abhishek =
Neeraj =
= 50 minutes
50
20×100
80
10×100
100
S190. Ans.(a)
Sol. Let Arunoday worked for 𝑥 minutes
2
5
5
5
𝑥
∴ 20 + 25 + 35 + 26 + 250 = 1
𝑥
= 25 minutes
250
𝑥 ≈ 91 minutes
∴ Required time = 91 – 5
= 86 minutes
= 10 minutes
Clearly on Tuesday, the efficiency of Neeraj is maximum. So he
should start the job so that the job is completed in the least
possible time.
S187. Ans.(b)
Sol. On Tuesday
Gaurav = 50 minutes
Arunoday =
150×100
5
= 100
11
89
100
1
1
+
500 25
3𝐶 2 × 5
𝐶2
8𝐶 3
𝐶1
8𝐶 3
15
= 28
15
= 56
Quantity II: boys = 10; girls = 10
10×9×8×7
10×9×8×7×6
Required ways = 10𝐶4 + 10𝐶5 = 4×3×2×1 + 5×4×3×2×1 = 462
Quantity I>Quantity II
= 125 min.
Neeraj = 10 min.
(Abhishek + Shailesh + Neeraj)’s 1 minute work = 10 + 2 + 25
= 37 units
Shailesh will work on this job for 7 minutes.
7×2
∴ Share of Shailesh =
× 875 = 49 Rs.
250
S189. Ans.(e)
Sol. On Tuesday —
Aman = 125 × 2 = 250 min.
Neeraj = 10 min.
Abhishek = 25 min.
Aman’s 50 min. work =
50
=
1
250
5
1
4
Remaining work = 1 − 5 = 5
4
let l and b
S195. Ans.(e)
Sol. Quantity I: Given a&b are roots of 𝑥 2 + 𝑥 − 6 = 0
Sum of roots, 𝑎 + 𝑏 = −1
Product of roots, 𝑎𝑏 = −6
1
1
𝑎+𝑏
−1
1
Required value = 𝑎 + 𝑏 = 𝑎𝑏 = −6 = 6
Quantity II: we know, (𝑎 + 𝑏)2 = 𝑎2 + 𝑏 2 + 2𝑎𝑏 = 20 +
2(8) = 36
𝑎 + 𝑏 = ±6
1
1
𝑎+𝑏
6
3
Required value = 𝑎 + 𝑏 = 𝑎𝑏 = 8 = 4 (𝑤ℎ𝑒𝑛 𝑎 + 𝑏 = +6)
1
+
1
𝑎+𝑏
−6
−3
No relation between quantity I & II.
10 25
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S194. Ans.(e)
Sol. Quantity I: perimeter = 2(l+b) = 48cm;
are 5x cm & 3x cm respectively.
8x = 24cm
X = 3cm
Therefore , Length= 15 cm and Breadth= 9 cm
Area = l × b = 15×9= 135 cm2
Quantity II: 135
Quantity I=Quantity II
Required value = 𝑎 + 𝑏 = 𝑎𝑏 = 8 = 4 (𝑤ℎ𝑒𝑛 𝑎 + 𝑏 = −6)
5
Required time = 1 5 1 = 5 7 minutes
70
Quantity II: required probability =
3𝐶 1 × 5
S193. Ans.(a)
Sol. Quantity I: boys = 12; girls = 20 – 12 = 8
12×11×10×9
8×7×6
Required ways = 12𝐶4 + 8𝐶5 = 4×3×2×1 + 3×2×1 = 551
4
= 21 21 minutes
S188. Ans.(d)
Sol. On Tuesday —
Abhishek = 25 min.
40
20
Quantity I>Quantity II
89
Remaining work = 1 − 100 = 100
50×100
× 60 = 100 𝑚𝑖𝑛
1
11
Shailesh =
120
Sol. Quantity I: required probability =
(Gaurav + Arunoday)’s 5 minutes work = 50 + 500 = 10 +
Required time =
200
S192. Ans.(a)
5
100
Required time =
Quantity I<Quantity II
Abhishek = 25 minutes
1
S191. Ans.(c)
Sol. Quantity I: Prabhas travels a distance of 100 km in 1 hour
when he stops for 10 min.
Actual travel time = 60 – 10 = 50 min
100
Speed of car = 50 × 60 = 120 𝑘𝑚𝑝ℎ
Quantity II: required time = 7+3 = 2 ℎ𝑜𝑢𝑟 = 120 𝑚𝑖𝑛
= 500 minutes
30
578
= 1 − 910
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S196. Ans.(c)
Sol. From I –
Let income of Sameer = 25x
96
So, income of Veer = 25𝑥 × 100 = 24𝑥
Let expenditure of Veer = 7y
So, expenditure of Sameer = 8y
3
Deepak spend th of his income.
S199. Ans.(b)
Sol. Let radius of circle = r cm
So, side of square = r + 3.5 cm
From I –
22
2 × × 𝑟 − 2𝑟 = 45
7
5
From II –
Saving of Sameer = 7000 Rs.
Saving of Veer = 7400
And, Income of Deepak is Rs. 1000 more than that of Sameer
From I & II –
(25𝑥−7000) 8𝑦
=
(24𝑥−7400) 7𝑦
r = 10.5 cm
side of square = 10.5 + 3.5 = 14 cm
Area of square = 196 cm2
Statement I alone is sufficient to give answer.
From II –
Let breadth of rectangle = 2x
So, radius of circle will be = 3x
ATQ –
22
×3𝑥
7
17x = 10200 ⇒ x = 600 Rs.
Income of Deepak = 25 × 600 + 1000 = 16000 𝑅𝑠.
2
Saving of Deepak = 5 × 16000 = 6400 Rs.
2×
S197. Ans.(b)
Sol. Let cost price = 100x
Marked price = 140x
From I –
75
140x ×
− 100𝑥 = 50
S200. Ans.(e)
Sol. From I –
Difference between speed of train A & B =10 meters/sec
And, length of train B = 240 meters
From I, we can’t determine
From II –
Train B cross pole in 8 sec
And train B cross train A in 12 sec
From II, we can’t determine
From I & II –
240
Speed of train B = 8 = 30 𝑚𝑒𝑡𝑒𝑟𝑠/𝑠𝑒𝑐
2(2𝑥+15)
Respective ratio of saving of Veer & Deepak = 7400 : 6400 =
37 : 32
So, Statement I & II together is sufficient to give answer of the
question.
100
5x = 50 ⇒ x = 10 Rs.
Cost price = 1000 Rs.
Statement I alone is sufficient.
From II –
6
90
(140x × 7 × 100 ) − 100𝑥 = 80
8x = 80 ⇒ x = 10 Rs.
Cost price = 1000 Rs.
Statement II alone is sufficient.
So, either statement I or II alone is sufficient to give answer of
the question.
(7−𝑥)(6−𝑥)
x = 3.5 cm
Radius of circle = 10.5 cm
side of square = 10.5 + 3.5 = 14 cm
Area of square = 196 cm2
So, either statement I or Statement II alone is sufficient.
Speed of train A = 30 − 10 = 20 meters/sec
Let length of train A = L meters
240+𝐿
So, (30 + 20) = 12
L = 600 − 240
L = 360 meters
So, Statement I and II both together sufficient to give answer
of the questions
S198. Ans.(a)
Sol. Given, number of green balls = 5
So, let total number of blue balls = x
So, number of red balls = (7 – x)
From I –
𝑥
7−𝑥
7
+ 12 = 12
12
So, we can’t determine value of x from statement I
From II –
𝑥 (𝑥−1)
3
=2
1
+ 12 ×11 = 6
2
2x – 14𝑥 + 42 = 22
2x2 – 14𝑥 + 20 = 0
2x2 – 10𝑥 − 4𝑥 + 20 = 0
2x(x – 5) −4 (𝑥 − 5) = 0
x = 2, 5
From II alone we can determine the difference between blue
& red balls in the bag.
So, only statement II alone is sufficient to give answer of the
question.
12 ×11
71
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S201. Ans.(e)
Sol.
√5776 - √1444 + √729 =43 + ?
76 – 38 +27 =43 + ?
?=65 -43 =22
S209. Ans.(d)
Sol.
1
21
5
1
41
55
+? = 1 −
20
21
+? = 21
1440
1330
=√81
=9
S212. Ans.(c)
Sol. 25×18+
4200
40
−
525
105
= 740 − ?
450+105−5=740 −?
?= 740−550
=190
S213. Ans.(d)
Sol.
3845+4380+2640 − 5965 = (?)2
(?)2=10865 – 5965
=4900
3
1
7
64
?=√4900
=70
S214. Ans.(b)
Sol.
400 ÷ 20 × 35 + 6666 ÷ 33+ ? = 1100
20× 35 + 202+? = 1100
?=1100-(700+202)
=1100- 902
=198
?
× 900 + 100 × 1050 = 100 × 3000
495 + 735 = 30 ×?
30 ×? = 1230
? = 41
S208. Ans.(b)
Sol. 73823 − 34156 + 4756 + 6758 − 9849 = 41499 −
160−?
41332 = 41339−?
?= 7
72
441
1
=√64 + 36 − 19
× 82 × 64 = 40
70
1
4
3840
S207. Ans.(a)
Sol. 55% 𝑜𝑓 900 + 70% 𝑜𝑓 1050 = ? % 𝑜𝑓 3000
100
20
Sol.√ 60 + 40 − 70
) ÷ (432 − 202 + 𝑜𝑓 21) × (82) =? 𝑜𝑓
×
=? −62
S211. Ans.(b)
1
16
49
?= 0
(16) ÷ (432 − 400 + 9) × (82) =?× 64
?=
88
7
1
84 × ×
S206. Ans.(e)
Sol.
5
×
1
S205. Ans.(b)
Sol.
2450 +3760 -3830 =6000 - ?
2380 =6000 - ?
?=6000 -2380 = 3620
64
2036
Sol. 84 × 4 ÷ 212 +? = 147 × 21 − 21
S204. Ans.(c)
Sol.
42.5×15 +37.5× 25= 1420 + ?
637.5+937.5 =1420 + ?
?= 1575 – 1420 = 155
𝑜𝑓25
3773
×
S210. Ans.(c)
S203. Ans.(a)
Sol.
1484÷28 + 1462÷34 -12×7= ?
?=53+43 -84 = 12
4
1331
14 =? −36
? = 50
S202. Ans.(a)
Sol. 78 ×26÷6 +1262= 1311 + (?)2
2028÷6+1262 =1311 +(?)2
338+1262 =1311+(?)2
(?)2=1600 -1311 =289
? =√289 =17
(5
5599
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S215. Ans.(b)
Sol.
28× 14.5+1680÷15+445=1000 -?
406+112+445=1000-?
963=1000-?
?=1000-963=37
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S216. Ans.(c)
Sol.
130
100
S223. Ans.(e)
Sol.
13
× 900
√1296 + √2025 + √1521– √? =
1250
× 1200 + 50 × 30 = ?
100
130 × 12 + 25 × 30 = ?
? = 1560 + 750
? = 2310
36 + 45 + 39 – √? = 117
√? = 120 – 117
?=9
S217. Ans.(a)
Sol.
S224. Ans.(b)
Sol.
56×240
350 + 14 + √? = (11)3
156
13
?
+ (3)2 × 40 = 100 × 600
√? = 1331– 350 – 960
√? = 21
? = 441
12 + 9 × 40 = ? × 6
372
? = 6 = 62
S225. Ans.(c)
Sol.
19×?
40
32 × 35 + √961 + 100 = 100 × 3305
S218. Ans.(a)
Sol.
1
680
√81 × 36 + 17 = ? + (512)3
1120 + 31 +
√2916 + 40 = ? + 8
? = 54+ 40 – 8 = 86
19×?
100
× 140 +
?=
640
16
?
100
× 1600 = 72 × 40
= 40
S220. Ans.(d)
Sol.
(17)2 + (22)2 + (8)2 + ? = 1750 − 820 + 2210
? + 289 + 484 + 64 = 1750 – 820 + 2210
? = 2303
S221. Ans.(a)
Sol.
308 + 672 –
40
100
×? +
80×355
100
2×?
2
S222. Ans.(b)
Sol.
8
= –619
S227. Ans.(c)
Sol.
(575 + 7511 – 2769) ÷ (76 × 1 + 675 – 342) = √?
= 5317 ÷ 409 = √?
⇒ ? = (13)2 = 169
S229. Ans.(d)
Sol.
= √(96) × 12 ÷ 18 + 26 – 9 = (65 – ? )% of 36
16
+ 25 × 42 – 100 × 400 = (32)2
= 1024 + 64 – 1050
? = 38 × 8 – 178
? = 126
73
S226. Ans.(b)
Sol.
1782 ÷ 54 + 456 – 2346 × 1 = ? × 3
⇒ 33 + 456 – 2346 = ? × 3
⇒ – 1857 = ? × 3
–1857
⇒ ?= 3
1
? = 1200
8
178+?
19
[(√3844 × 9) ÷ (27)3 ] × 23 =?2 + 337
⇒ [(62 × 3) ÷ 3] × 23 =?2 + 337
⇒ 1426 – 337 = ?²
⇒ ? = √1089
= 33
480×5
178+?
= 1322 – 1151
S228. Ans.(a)
Sol.
= (28)2
980 + 284 – 784 = 5
?=
= 1322
? = 900
16 × 140 + 16 × ? = 72 × 40
2240 + 16 × ? = 2880
?=
100
100
171×100
S219. Ans.(e)
Sol.
1600
19×?
⇒9=
(65 –?)
100
× 36 ⇒ (65 – ? ) =
⇒ ? = 65 – 25 = 40
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36
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S230. Ans.(a)
Sol.
12 × √225 + 1212 – (1053 ÷ 9) = ?
⇒ 1392 – (117) = ?
⇒ ? = 1275
S238. Ans.(d)
Sol.
35.99 × 4.98 − 1199.99 ÷ 7.99 =?
36 × 5 −
12
+? =
=?
S239. Ans.(b)
Sol.
?2 + 60% 𝑜𝑓 239.99 = 55% 𝑜𝑓 320.02 + 3.98
60
S232. Ans.(d)
Sol.
√(123.09 + 465.05) ÷ 11.99+? = 240.02 ÷ 1.989
123+465
8
? = 180 − 150
? = 30
S231. Ans.(b)
Sol.
42
28
× 350 − 100 × 400 =?
100
147 − 112 =?
? = 35
√
1200
55
?2 + 100 × 240 = 100 × 320 + 4
?2 + 144 = 176 + 4
?2 = 180 − 144
?= 6
240
2
S240. Ans.(c)
Sol.
524.90 + 125.05 =? × 9.99
525 + 125 =?× 10
√49+? = 120
? = 113
S233. Ans.(e)
Sol.
(15.99)2 − 14.04 × 8.99+? = 154.999
162 − 14 × 9+? = 155
? = 155 + 126 − 256
? =25
650
? = 10
? = 65
S241. Ans.(e)
Sol.
√144.04 × 15% 𝑜𝑓 120.09 =? −54.99 × 3.03
S234. Ans.(a)
Sol.
62
20
× 250 −
× 105−? = 110
100
100
155 − 21 − 110 =?
? = 24
15
√144 × 100 × 120 =? −55 × 3
12 × 18 =? −165
? = 216 + 165
? = 381
S235. Ans.(c)
Sol.
45% 𝑜𝑓 220.09 + 30% 𝑜𝑓 160.06 =?2 + 2.99
45
30
× 220 + 100 × 160 =?2 + 3
100
99 + 48 − 3 =?2
? = 12
S242. Ans.(a)
Sol.
13.03 × 7+? = 30.03% 𝑜𝑓 349.99
13 × 7+? =
30
100
× 350
91+? = 105
? = 14
S236. Ans.(a)
Sol.
1229.99 + 2120.09 − 3049.987 =?
1230 + 2120 − 3050 =?
? = 300
S237. Ans.(e)
Sol.
√√(99.99 + 104.99 × 5 =?÷ 8.989
√√100 + 105 × 5 = ?
9
√√625 = ?
9
? = 45
74
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S243. Ans.(d)
Sol.
32% 𝑜𝑓 600.02 − 19.99% 𝑜𝑓 400.04+? = 859.99 ÷ 2
32
100
20
860
× 600 − 100 × 400+? = 2
192 − 80+? = 430
? = 318
S244. Ans.(b)
Sol.
141
279.89
+
− √? = 10.01
S249. Ans.(a)
Sol.
Wrong no. = 120
118 + 7 = 125
125 + 11 = 136
136 + 13 = 149
149 + 17 = 166
166 + 19 = 185
185 + 23 = 208
20.09
39.99
140
280
20
+
40
− √? = 10
√? = 7 + 7 − 10
? = 16
S245. Ans.(c)
Sol.
8.98 × 60.02 − 19.992 + 10.01% 𝑜𝑓 130.09 =?
10
9 × 60 − 202 + 100 × 130 =?
540 − 400 + 13 =?
? = 153
S250. Ans.(b)
Sol.
Wrong no. = 214
205 + 23 = 213
213 − 33 = 186
186 + 43 = 250
250 − 53 = 125
125 + 63 = 341
341 − 73 = −2
S246. Ans.(d)
Sol.
Wrong no. = 22
28 × 0.5 = 14
14 × 1 = 14
14 × 1.5 = 21
21 × 2 = 42
42 × 2.5 = 105
105 × 3 = 315
S251. Ans.(b)
Sol. Wrong number = 810
Pattern of series –
S247. Ans.(e)
Sol.
Wrong no. = 47
5 + (12 + 1) = 7
7 + (22 + 2) = 13
13 + (32 + 3) = 25
25 + (42 + 4) = 45
45 + (52 + 5) = 75
75 + (62 + 6) = 117
S252. Ans.(d)
Sol. Wrong number = 350
Pattern of series –
S248. Ans.(d)
Sol.
Wrong no. = 2400
288000 ÷ 12 = 24000
24000 ÷ 10 = 2400
2400 ÷ 8 = 300
300 ÷ 6 = 50
50 ÷ 4 = 12.5
12.5 ÷ 2 = 6.25
S253. Ans.(a)
Sol. Wrong number = 646
Pattern of series –
75
814
820
832
+ 12
+6
868
+ 36
+ 144
×4
×3
×2
1732
1012
+ 4320
+ 720
×5
6052
×6
So, there should be 814 in place of 810.
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348
1024
832
-6 7 6
+484
(2 6 )²
(2 2 )²
508
+196
-3 2 4
(1 4 )²
(1 8 )²
604
704
640
-1 0 0
+36
(1 0 )²
(6 )²
So, there should be 348 in place of 350.
190
210
266
+ 56
+ 20
+ 36
358
+ 92
+ 36
+ 128
+ 36
+ 200
+ 164
+ 36
850
650
486
+ 36
So, there should be 650 in place of 646.
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S254. Ans.(e)
S259. Ans.(a)
Sol. Wrong number = 1
Pattern of series –
Sol. Wrong number = 15
Pattern of series –
17
50
160
370
+ 11 0
+33
+210
+160
1904
+696
+499
+339
+129
+100
+77
1208
709
+197
So, there should be 2 in place of 1.
+29
+23
+31
+37
So, there should be 17 in place of 15.
S260. Ans.(b)
Sol. Wrong number = 1990
Pattern of series –
S255. Ans.(b)
Sol. Wrong number = 990
Pattern of series –
120
170
251
+81
+50
367
+4
+198
+43
+39
+35
+31
+155
+ 11 6
722
522
+245
+47
So, there should be 1992 in place of 1990.
+4
+4
+4
So, there should be 965 in place of 990.
S261. Ans.(a)
Sol.
I:
2𝑥 2 + 𝑥 − 6 = 0
2𝑥 2 + 4𝑥 − 3𝑥 − 6 = 0
2𝑥(𝑥 + 2) − 3(𝑥 + 2) = 0
(2𝑥 − 3) (𝑥 + 2) = 0
𝑥 = 1.5, − 2
II:
𝑦 2 + 6𝑦 + 9 = 0
𝑦 2 + 3𝑦 + 3𝑦 + 9 = 0
𝑦(𝑦 + 3) + 3(𝑦 + 3) = 0
(𝑦 + 3)(𝑦 + 3) = 0
𝑦 = −3, −3
So, 𝑥 > 𝑦
S256. Ans.(b)
Sol. Wrong number = 1090
Pattern of series –
So, there should be 1080 in place of 1090.
S257. Ans.(e)
Sol. Wrong number = 110
Pattern of series –
S262. Ans.(d)
Sol.
I:
𝑥 2 − 4𝑥 + 4 = 0
𝑥 2 − 2𝑥 − 2𝑥 + 4 = 0
𝑥(𝑥 − 2) − 2(𝑥 − 2) = 0
(𝑥 − 2)(𝑥 − 2) = 0
𝑥 = 2, 2
II:
𝑦 2 − 10𝑦 + 16 = 0
𝑦 2 − 8𝑦 − 2𝑦 + 16 = 0
𝑦(𝑦 − 8) − 2(𝑦 − 8) = 0
(𝑦 − 8)(𝑦 − 2) = 0
𝑦 = 8, 2
So, 𝑥 ≤ 𝑦
So, there should be 113 in place of 110.
S258. Ans.(d)
Sol. Wrong number = 255
Pattern of series –
So, there should be 256 in place of 255.
76
965
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S263. Ans.(e)
Sol.
I:
2𝑥 2 + 7𝑥 + 6 = 0
2𝑥 2 + 3𝑥 + 4𝑥 + 6 = 0
𝑥(2𝑥 + 3) + 2(2𝑥 + 3) = 0
(2𝑥 + 3)(𝑥 + 2) = 0
S266. Ans.(a)
Sol.
I:
4𝑥 2 − 20𝑥 + 25 = 0
4𝑥 2 − 10𝑥 − 10𝑥 + 25 = 0
2𝑥(2𝑥 − 5) − 5(2𝑥 − 5) = 0
(2𝑥 − 5)(2𝑥 − 5) = 0
5 5
3
𝑥 = − 2 , −2
𝑥 = 2,2
II:
3𝑦 2 + 11𝑦 + 10 = 0
3𝑦 2 + 6𝑦 + 5𝑦 + 10 = 0
3𝑦(𝑦 + 2) + 5(𝑦 + 2) = 0
(3𝑦 + 5)(𝑦 + 2) = 0
II:
5𝑦 2 − 6𝑦 − 8 = 0
5𝑦 2 − 10𝑦 + 4𝑦 − 8 = 0
5𝑦(𝑦 − 2) + 4(𝑦 − 2) = 0
(5𝑦 + 4)(𝑦 − 2) = 0
4
5
𝑦 = 2, − 5
𝑦 = − 3 , −2
So, no relation can be established.
So, 𝑥 > 𝑦
S264. Ans.(d)
Sol.
I:
𝑥 2 − 2𝑥 − 24 = 0
𝑥 2 − 6𝑥 + 4𝑥 − 24 = 0
𝑥(𝑥 − 6) + 4(𝑥 − 6) = 0
(𝑥 − 6)(𝑥 + 4) = 0
𝑥 = −4, 6
II:
𝑦 2 − 12𝑦 + 36 = 0
𝑦 2 − 6𝑦 − 6𝑦 + 36 = 0
𝑦(𝑦 − 6) − 6(𝑦 − 6) = 0
(𝑦 − 6)(𝑦 − 6) = 0
𝑦 = 6, 6
So, 𝑥 ≤ 𝑦
S267. Ans.(b)
Sol.
I:
𝑥 2 − 2𝑥 − 15 = 0
𝑥 2 − 5𝑥 + 3𝑥 − 15 = 0
𝑥(𝑥 − 5) + 3(𝑥 − 5) = 0
(𝑥 + 3) (𝑥 − 5) = 0
𝑥 = −3, 5
II:
𝑦 2 − 15𝑦 + 56 = 0
𝑦 2 − 8𝑦 − 7𝑦 + 56 = 0
𝑦(𝑦 − 8) − 7(𝑦 − 8) = 0
(𝑦 − 7)(𝑦 − 8) = 0
𝑦 = 7, 8
So, 𝑥 < 𝑦
S265. Ans.(a)
Sol.
I:
4𝑥 2 + 11𝑥 + 6 = 0
4𝑥 2 + 8𝑥 + 3𝑥 + 6 = 0
4𝑥(𝑥 + 2) + 3(𝑥 + 2) = 0
(4𝑥 + 3)(𝑥 + 2) = 0
S268. Ans.(e)
Sol.
I:
10𝑥 2 + 19𝑥 + 7 = 0
10𝑥 2 + 14𝑥 + 5𝑥 + 7 = 0
2𝑥(5𝑥 + 7) + 1(5𝑥 + 7) = 0
(2𝑥 + 1)(5𝑥 + 7) = 0
1
7
2
5
𝑥 = − , −2
𝑥 = − ,−
II:
𝑦 2 + 10𝑦 + 25 = 0
𝑦 2 + 5𝑦 + 5𝑦 + 25 = 0
𝑦(𝑦 + 5) + 5(𝑦 + 5) = 0
(𝑦 + 5)(𝑦 + 5) = 0
𝑦 = −5, −5
So, 𝑥 > 𝑦
II:
5𝑦 2 + 16𝑦 + 12 = 0
5𝑦 2 + 6𝑦 + 10𝑦 + 12 = 0
𝑦(5𝑦 + 6) + 2(5𝑦 + 6) = 0
(𝑦 + 2)(5𝑦 + 6) = 0
3
4
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6
𝑦 = −2, − 5
So, no relation can be established.
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S269. Ans.(a)
Sol.
I:
𝑥 2 − 20𝑥 + 75 = 0
𝑥 2 − 15𝑥 − 5𝑥 + 75 = 0
𝑥(𝑥 − 15) − 5(𝑥 − 15) = 0
(𝑥 − 5)(𝑥 − 15) = 0
𝑥 = 5, 15
II:
𝑦 2 + 19𝑦 + 84 = 0
𝑦 2 + 12𝑦 + 7𝑦 + 84 = 0
𝑦(𝑦 + 12) + 7(𝑦 + 12) = 0
(𝑦 + 12)(𝑦 + 7) = 0
𝑦 = −12, −7
So, 𝑥 > 𝑦
S272. Ans.(c)
Sol.
I:
𝑥 2 − 18𝑥 + 56 = 0
𝑥 2 − 14𝑥 − 4𝑥 + 56 = 0
𝑥(𝑥 − 14) − 4(𝑥 − 14) = 0
(𝑥 − 4)(𝑥 − 14) = 0
𝑥 = 4, 14
II:
𝑦 2 + 4𝑦 − 32 = 0
𝑦 2 + 8𝑦 − 4𝑦 − 32 = 0
𝑦(𝑦 + 8) − 4(𝑦 − 8) = 0
(𝑦 − 4)(𝑦 + 8) = 0
𝑦 = −8, 4
So, 𝑥 ≥ 𝑦
S273. Ans.(d)
Sol.
I:
𝑥 2 + 14𝑥 − 72 = 0
𝑥 2 + 18𝑥 − 4𝑥 − 72 = 0
𝑥(𝑥 + 18) − 4(𝑥 + 18) = 0
(𝑥 + 18)(𝑥 − 4) = 0
𝑥 = −18, 4
II:
𝑦 2 − 13𝑦 + 36 = 0
𝑦 2 − 9𝑦 − 4𝑦 + 36 = 0
𝑦(𝑦 − 9) − 4(𝑦 − 9) = 0
(𝑦 − 4)(𝑦 − 9) = 0
𝑦 = 4, 9
So, 𝑥 ≤ 𝑦
S270. Ans.(e)
Sol.
I:
𝑥 2 − 9𝑥 − 22 = 0
𝑥 2 − 11𝑥 + 2𝑥 − 22 = 0
𝑥(𝑥 − 11) + 2(2𝑥 − 11) = 0
(𝑥 + 2)(𝑥 − 11) = 0
𝑥 = −2, 11
II:
𝑦 2 − 17𝑦 + 66 = 0
𝑦 2 − 11𝑦 − 6𝑦 + 66 = 0
𝑦(𝑦 − 11) − 6(𝑦 − 11) = 0
(𝑦 − 11)(𝑦 − 6) = 0
𝑦 = 6, 11
So, no relation can be established.
S274. Ans.(d)
Sol.
I:
𝑥 2 − 92 = 122
𝑥 2 = 144 + 81
𝑥 2 = 225
𝑥 = 15, −15
II:
𝑦 3 = 3375
𝑦 = 15
So, 𝑥 ≤ 𝑦
S271. Ans.(b)
Sol.
I:
4𝑥 2 + 19𝑥 + 15 = 0
4𝑥 2 + 15𝑥 + 4𝑥 + 15 = 0
𝑥(4𝑥 + 15) + 1(4𝑥 + 15) = 0
(4𝑥 + 15)(𝑥 + 1) = 0
𝑥 = −1, −15
II:
8𝑦 2 + 10𝑦 + 3 = 0
8𝑦 2 + 6𝑦 + 4𝑦 + 3 = 0
2𝑦(4𝑦 + 3) + 1(4𝑦 + 3) = 0
(4𝑦 + 3)(2𝑦 + 1) = 0
3
S275. Ans.(b)
Sol. I:
5
𝑥2
28
5 3
1
So, 𝑥 < 𝑦
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𝑥2
7
28
𝑥 2− 2 = 7
𝑥=4
II:
11𝑦 + (7 × 6) = 97
11𝑦 + 42 = 97
11𝑦 = 55
𝑦 =5
So, 𝑥 < 𝑦
𝑦 = −4,−2
78
3
=
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S276. Ans.(a)
Sol.
I. 𝑥 2 − 22𝑥 + 72 = 0
x2 -18x -4x +72=0
x(x-18)-4(x-18)=0
(x-18)(x-4)=0
x = 18, 4
II. 𝑦 2 + 11𝑦 + 30 = 0
y2+5y +6y +30=0
y(y+5) +6(y+5)=0
(y+5)(y+6)=0
y= -5,-6
So, x>y
S280. Ans.(c)
Sol.
I. 𝑥 2 + 4𝑥 − 12 = 0
x2+6x-2x-12=0
x(x+6)-2(x+6)=0
(x+6) (x-2)=0
x=-6, 2
II. 𝑦 2 − 9𝑦 + 20 = 0
y2-5y-4y+20=0
y(y-5)-4(y-5)=0
(y-5) (y-4)=0
y=5, 4
𝑠𝑜, 𝑦 > 𝑥
S277. Ans.(e)
Sol.
I. 𝑥 2 − 23𝑥 + 120 = 0
x2 -15x -8x +120=0
x(x-15)- 8(x-15)=0
(x-15)(x-8)=0
x = 15, 8
II. 𝑦 2 − 17𝑦 + 70 = 0
y2-10y -7y +70=0
y(y-10) -7(y-10)=0
(y-7)(y-10)=0
y= 10, 7
So, no relation can be established
S281. Ans.(e)
Sol.
I. 𝑥 2 − 25𝑥 + 100 = 0
x2 -20x -5x +100=0
x(x-20)-5(x-20)=0
(x-20) (x-5)=0
x =20,5
II. 𝑦 2 − 27𝑦 + 110 = 0
y2-22y -5y +110=0
y(y-22) -5(y-22)=0
(y-22)(y-5)=0
y=22,5
So, no relation can be established between x and y.
S278. Ans.(b)
Sol.
I. 𝑥 2 − 15𝑥 + 54 = 0
x2-9x-6x+54=0
x(x-9)-6(x-9)=0
(x-9)(x-6)=0
x= 9, 6
II. 𝑦 2 + 10𝑦 − 96 = 0
y2+16y-6y-96 =0
y(y+16)-6(y+16)=0
(y+16)(y-6)=0
y = 6,−16
So, 𝑥 ≥ 𝑦
S282. Ans.(d)
Sol.
I. 𝑥 2 = 289
x =√289
x =17, −17
II. 𝑦 = √289
y =17
So, x ≤ y
S279. Ans.(b)
Sol.
I.x3 +440=2168
x3=2168-440
x3=1728
x =12
II. 𝑦 2 − 23 =121
y2=121+23
y2=144
y =12, −12
So, x ≥ y
79
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S287. Ans.(b)
Sol.
I. 𝑥 2 − 14𝑥 + 45 = 0
𝑥 2 − 9𝑥 − 5𝑥 + 45 = 0
𝑥(𝑥 − 9) − 5(𝑥 − 9) = 0
(𝑥 − 9)(𝑥 − 5) = 0
𝑥 = 9, 5
II. 𝑦 2 + 2𝑦 − 35 = 0
𝑦 2 + 7𝑦 − 5𝑦 − 35 = 0
𝑦(𝑦 + 7) − 5(𝑦 + 7) = 0
(𝑦 − 5)(𝑦 + 7) = 0
𝑦 = 5, −7
So, 𝑥 ≥ 𝑦
S283. Ans.(d)
Sol.
I. 𝑥 2 + 12𝑥 + 32 = 0
x2+8x+4x+32=0
x(x+8)+4(x+8)=0
(x+8)(x+4)=0
x= −8, −4
II. 𝑦 2 + 7𝑦 + 12 = 0
y2+ 3y+4y+12 =0
y(y+3)+4(y+3)=0
(y+4)(y+3)=0
y = −4, −3
So, 𝑦 ≥ 𝑥
S284. Ans.(a)
Sol.
I. 3𝑥² + 16𝑥 + 20 = 0
3x2 +6x +10x +20 =0
3x(x+2) +10(x+2)=0
(3x+10)(x+2)=0
10
x= −2, −
3
II. 𝑦² + 14𝑦 + 48 =0
y2 +8y +6y +48 =0
y(y +8)+6(y+8)=0
(y+6)(y+8)=0
y =−6, −8
So, x > y
𝐒𝟐85. Ans.(e)
Sol.
I. 𝑥 2 + 𝑥 − 72 = 0
x2+ 9x-8x-72=0
x(x+9)-8(x+9)=0
(x+9)(x-8)=0
x = 8,−9
II. 𝑦 2 + 13𝑦 + 42=0
y2 +6y+7y+42 =0
y(y+6)+7(y+6)=0
(y+6)(y+7)=0
y = -6,-7
So, no relation can be established between x and y.
S286. Ans.(c)
Sol.
I. 𝑥 2 + 5𝑥 + 6 = 0
𝑥 2 + 3𝑥 + 2𝑥 + 6 = 0
𝑥(𝑥 + 3) + 2(𝑥 + 3) = 0
(𝑥 + 3)(𝑥 + 2) = 0
𝑥 = −2, −3
II. 𝑦 2 − 9𝑦 + 14 = 0
𝑦 2 − 7𝑦 − 2𝑦 + 14 = 0
𝑦(𝑦 − 7) − 2(𝑦 − 7) = 0
(𝑦 − 2)(𝑦 − 7) = 0
𝑦 = 2, 7
So, y>x
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S288. Ans.(e)
Sol.
I. 𝑥 2 + 11𝑥 + 18 = 0
𝑥 2 + 9𝑥 + 2𝑥 + 18 = 0
𝑥(𝑥 + 9) + 2(𝑥 + 9) = 0
(𝑥 + 9)(𝑥 + 2) = 0
𝑥 = −2, −9
II. 𝑦 2 + 6𝑦 + 8 = 0
𝑦 2 + 4𝑦 + 2𝑦 + 8 = 0
𝑦(𝑦 + 4) + 2(𝑦 + 4) = 0
(𝑦 + 4)(𝑦 + 2) = 0
𝑦 = −4, −2
So, no relation can be established
S289. Ans.(c)
Sol.
I. 𝑥² + 5𝑥 + 6 = 0
𝑥 2 + 3𝑥 + 2𝑥 + 6 = 0
𝑥(𝑥 + 3) + 2(𝑥 + 3) = 0
(𝑥 + 3)(𝑥 + 2) = 0
𝑥 = −3, −2
II. 𝑦 2 – 15𝑦 = 16
𝑦 2 − 15𝑦 − 16 = 0
𝑦 2 − 16𝑦 + 𝑦 − 16 = 0
𝑦(𝑦 − 16) + 1(𝑦 − 16) = 0
(𝑦 + 1)(𝑦 − 16) = 0
𝑦 = −1, 16
Clearly, 𝑥 < 𝑦
S290. Ans.(a)
Sol.
Multiplying II by 3 and subtracting II from I, we get,
𝑦 = −1 𝑎𝑛𝑑 𝑥 = 3
So, 𝑥 > 𝑦
S291. Ans.(b)
Sol. Required percentage =
(22+23)−10
22+23
× 100
≈ 78%
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S292. Ans.(c)
S299. Ans.(b)
Sol. Population which is literate but ungraduated from
1
Sol. Required average = 3 × (12 + 15 + 18)% × 22500
city A = 10000 ×
15
= 100 × 22500
60
100
= 6000
6000
Required percentage = 2400 × 100 = 250%
= 3375
S293. Ans.(d)
S300. Ans.(c)
23−12
Sol. Required no. of passenger = 22500 × 100 = 2475
Sol. Graduate male from city C =
S294. Ans.(d)
Literate but ungraduated from city B = 6000 × 100 = 3600
25600
16
× 9 = 14400
60
10
7
23
5
Sol. Required ratio = 22500 × 100 × 15 : 22500 × 100 × 23
= 14: 15
Required difference = 14400 − 3600 = 10800
S301. Ans.(b)
2200
Sol. Passenger travelling to Rewari = 22500 ×
18
100
= 4050
15
Passenger travelling to Panipat = 22500 × 100 = 3375
4
1200
Selling price of per kg sugar = 40 = 𝑅𝑠. 30
Required difference = 30 − 25 = 𝑅𝑠. 5 less
Required difference = 3375 × 75 × − 4050 × 75
S302. Ans.(e)
= 75 × (4500 − 4050)
= 75 × 450
= 33750 Rs.
Sol (96-100): For city C
Sol. selling price of one kg wheat =
3
Required average selling price =
2
20×3+10×2
80
= 5
3+2
S303. Ans.(b)
900
Sol. Required percentage = 2200−600 × 100 = 56.25%
1
Illiterate population = 96000 × = 32000
3
40
= 𝑅𝑠. 20
= Rs. 16 per kg
Literate population of city C = 96000 × 3 = 64000
100
45
Selling price of one kg salt = 60 = 𝑅𝑠. 10
6000
Graduate population = 64000 ×
900
600
Total population of city C = 6.25 × 100 = 96000
= 25600
S304. Ans.(b)
For city B
Total population = 16000
Literate population = 6000
Illiterate population = 16000 − 6000 = 10000
Sol. selling price of a kg pulse =
3750
50
= 𝑅𝑠. 75
Profit earned on selling of one kg pulse = 75 − 60 =
𝑅𝑠. 15
Total profit = 15 × 40 = 𝑅𝑠. 600
40
Graduate population = 6000 × 100 = 2400
For city A
Total population = 22000
S305. Ans.(e)
1
Sol. Required average quantity = 3 × (55 + 50 + 45)
5
Literate population = 22000 × 11 = 10000
150
= 3
Illiterate population = 22000 − 10000 = 12000
40
= 50 kg
Graduate population = 10000 × 100 = 4000
S306. Ans.(a)
Sol.
No. of male student playing Hockey of college L
S296. Ans.(c)
6000
Sol. Required percentage = 12000 × 100 = 50%
8
= 450 × 9 = 400
S297. Ans.(d)
Sol. Required ratio = 25600: 16000
= 8:5
Average no. of student playing Hockey of college M & O
=
S298. Ans.(a)
Sol. Required difference = 32000 − 2400 = 29600
81
100
Sol. Cost price of per kg rice = 55 × 160 = 𝑅𝑠. 25
S295. Ans.(a)
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400+500
2
= 450
400
8
Required percentage = 450 × 100 = 88 9 %
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S307. Ans.(c)
Sol.
Student who left playing Cricket of college N
S311. Ans.(b)
Sol. USA has lowest mortality rate, which is 3.14%
1
= 350 × 7 = 50
S312. Ans.(d)
Total student playing Football of college N
= 450 + 50 = 500
Required ratio =
500+300
500+300
Sol. Required % =
=1∶ 1
(250+450+500)
3
= 400
Average no. of student playing Football of college K, L and
M
=
400+350+300
3
= 350
650
× 100 = 7
= 15 : 7
S314. Ans.(a)
Sol. Required % =
4000+11000+17500+15000
80000
× 100 =
59.375%
5
9
13
4
100000
Mortality rate in china is 5%.
New number of total deaths = 100000×
5
100
= 5000
S316. Ans.(e)
11
Sol. Price of a one kg sugar = 84 × 21 = 𝑅𝑠 44
10
Price of one kg of salt = 840 × 21 = 𝑅𝑠40
%
S310. Ans.(d)
Sol.
Total student in college K in 2014 = 400 + 500 + 250 =
1150
Total student in college K in 2015
120
= 1150 × 100 = 1380
Student playing Football of college K in 2015
Required difference = (20 × 44 – 15 × 40)
= 880 – 600
= 𝑅𝑠. 280
S317. Ans.(a)
Sol. Price of one kg of tea =
Price of one kg of rice =
5
= 1380 × 10
Required % =
= 690
400+690
Required average =
2
=
80000
×100
×100
Sol. New total confirmed cases in china = 80000 × =
S309. Ans.(e)
Sol.
Total no. of student playing Cricket of college L and M
together
= 400 + 300 = 700
Total no. of student playing Hockey of college K and M
together
= 250 + 400 = 650
700–650
15000
Sol. Required ratio = 140000
4000
S315. Ans.(e)
Required difference = 400 – 350 = 50
Required percentage =
× 100 = 1900%
17500
S313. Ans.(c)
S308. Ans.(b)
Sol.
Average no. of student playing Hockey of college K, L and
O
=
350000−17500
50−50
50
900
1500
30
18
= 𝑅𝑠50
= 𝑅𝑠50
× 100 = 0%
S318. Ans.(d)
1090
2
63×12
18
= 545
Sol. Required ratio = 42×25 = 25
Solution (111-115):
ATQ,
4000
Mortality rate for China = 80000 × 100 = 5%
S319. Ans.(b)
20+15
1
Required%= 30+12 × 100 = 83 3 %
11000
Mortality rate for USA = 350000 × 100 = 3.14%
17500
Mortality rate for Italy = 130000 × 100 = 13.46%
Mortality rate for Spain =
82
15000
140000
× 100 = 10.71%
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S320. Ans.(b)
Sol. Required sum = (56 × 15) + (32 × 30) + (40 × 25)
= 2800 Rs.
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S321. Ans.(c)
Sol.
Let length of train A = l metres.
And let speed of train A = S m/s.
ATQ,
Speed of train B =
S325. Ans.(b)
Sol.
let speed of boat in still water and speed of Stream be P
and Q kmph respectively.
ATQ,
450+150
Speed of train A, S =
𝑙+230
450+𝑙
450+𝑙
160
= 25 −
S326. Ans.(c)
Sol.
Given distance between P and Q is 900 km.
𝑙+230
l=350 metres.
= 8 𝑘𝑚𝑝ℎ (Upstream Speed)
…(ii)
160
On solving (i) & (ii):
𝑙+450
5
P+Q = 16 𝑘𝑚𝑝ℎ (Downstream Speed)
ATQ, Downstream Speed, X-4 = P+Q
So, X = 16+4 = 20.
…(i)
29
Now, 25 − 𝑆 = 160
𝑆 = 25 −
40
P-Q =
24
= 25 m/s
29
So, speed of train A =
350+230
= 20 m/s.
Required time =
speed of car B =
29
20
S322. Ans.(e)
Sol. Let the speed of train A and train B be 17X m/s and
13X m/s respectively.
And let the length of train B = Y meter
950+𝑌
ATQ, 17𝑋−13𝑋 = 16
Y = 64X -950,
So, length can’t be determined with given data.
900
(𝑋+4)
9
900
5
×2= 𝑋 ×2
 9X = 5 (X+4)
 4X = 20
So, speed of car B =
900
(5+4)
=100 kmph.
9
Required distance= 100 × = 450 km
2
S327. Ans.(b)
Sol.
Now, let speed of the boat in still water and the speed of
the stream be a km/hr. & b km/hr. respectively.
So, upstream speed of boat = (𝑎 − 𝑏) km/hr.
ATQ,
𝑎 − 𝑏 = 15
5
3𝐿 = 144 × 18 × 30
L = 400 m
Length of train = 800 m
∴ Length of other train = 2 × 800 = 1600 m
5
ATQ,
Car B started from P at 6:00am
and car A started from P at 8:00 am
They both met at 10:30 am i.e.
X = 5 hours
S323. Ans.(d)
Sol.
Let length of train = 2L m
Length of tunnel = L m
ATQ,
60
60% of speed = 144 × 18 × 100 = 24 m/sec.
120
∴ (1600 + 800) = 24 × time
∴ time = 100 sec.
Required time= (𝑎−𝑏)
S324. Ans.(b)
Sol.
Let us assume the original speed of Deepak be 4x km/hr
and original time taken by Deepak be T hr.
ATQ, decreased speed of Deepak = 3x km/hr,
=8 hr.
=
24
And increased time of Deepak = (T+60)
= (𝑇 + 0.40) hours
So, 4𝑥 × 𝑇 = 3𝑥 × (𝑇 + 0.4)
T = 1.2hour = 72 minutes
83
km/h.
Speed of car A = 𝑋 km/h.
350+50
= 20 sec.
900
(𝑋+4)
900
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120
15
S328. Ans.(c)
Sol.
Let the distance between Amit’s home and his office is D
km.
𝐷
𝐷
2𝐷
ATQ, 30 + 𝑋 = 33
X = 36.67 km/hr
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S329. Ans.(a)
Sol.
Time taken by X = 8 hr.
Time taken by Y = 7 hr.
S332. Ans.(b)
Sol.
Let speed of stream = r km/h
A/q,
(8 − 𝑟) × 5 = (8 + 𝑟) × 3
⇒ 40 – 5r = 24 + 3r
16
⇒ 𝑟 = 8 = 2 𝑘𝑚/ℎ
∴ time taken to cross each other
56
S333. Ans.(b)
Sol.
Let total distance = d
11
= 15 = 3 15 hr.
= 3 hr 44 min.
∴ Required time to cross = 11 : 44 am
∴ Average speed = 𝑑
S330. Ans.(b)
Sol.
Let initial speed of the car = s kmph.
And initial time taken by the car to cover the distance = t
hours.
So, Total Distance = 𝑠 × 𝑡 km.
ATQ,
(𝑠 − 9)(𝑡 + 2) = (𝑠 + 5) (𝑡 −
48
60
)
s-5t = 5 …..(i)
and,
st = (s-9) (t+2)
2s-9t = 18 …….(ii)
From eq(i) & eq(ii)
t=8 hours
and s= 45 kmph
so, required distance = 45 × 8 = 360 𝑘𝑚.
1100
700
S334. Ans.(a)
Sol.
Let the total distance = x km
x
x
90
+
=
12– 4 12 + 4 60
x
x
+
= 1.5
8 16
3x = 1.5 × 16
x = 8km
2
⇒ x = 7t
Let t = 7y = time taken on train
x = 2y = time taken on car
t−x = 5y = time taken on cycle.
Required Ratio →
60 × 2y : 4 × 5y : 20 × 7y
6
: 1
: 7
× 7𝑥
= 11x km/hr.
Hence, speed of boat in still water =
𝑑
+
24 48
S335. Ans.(d)
Sol.
Let one side time taken = t hour
Time taken by car = x hour
ATQ,
60x + 4(t – x) = 20 × t
S331. Ans.(b)
Sol.
Let upstream speed of a boat be 7x km/hr.
So, downstream speed of a boat =
= 16 km/h
𝑑
7𝑥+11𝑥
2
= 9x km/hr.
And, speed of stream = 11𝑥 − 9𝑥
= 2x km/hr.
ATQ,
2𝑥 = 8
𝑥=4
176
70
Required time = 11𝑥 + 7𝑥
16
10
= 𝑥 +𝑥
26
= 𝑥
= 6.5 hours
84
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S336. Ans.(c)
Sol.
In both conical shape volume will be same.
Let base radius of cone is R cm
So, height of the cone = 2R cm.
ATQ,
4
3
S341. Ans.(d)
Sol.
Total number of cases when two dices are rolled
simultaneously=36
total cases of getting same number on both the
dices=(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6
6
5
required probability=1 − 36 = 6
1
𝜋(16)3 = 2 × 𝜋𝑅2 × 2𝑅
3
𝑅3 = 163
R =16
Required height = 2R =32 cm.
S342. Ans.(d)
Sol.
4
Volume of sphere = 𝜋𝑅3 (R → Radius)
3
S337. Ans.(d)
Sol. According to question the first place of the three-letter
word will be fix & will be filled by S only.
So, rest two letter will be selected from the rest 6 letter of
word STRANGE.
So, Number of possible ways = 6 × 5 = 30
S338. Ans.(b)
Sol. total outcomes = 62 = 36
Favorable outcomes = when sum is 2, 3, 4, 8, 9, 10
(1,1) (1,2) (1,3) (2,1) (2,2) (2,6) (3,1) (3,5) (3,6) (4,4)
(4,5) (4,6) (5,3) (5,4) (5,5) (6,2) (6,3) (6,4)
Required probability =
18
36
=
2
1
2
× 𝑎 × 𝑎 = 2048 m2
a2 = 4096.
a=64 m.
so, its hypotenuse = 64√2 m.
7.5×7.5×7.5
S345. Ans.(a)
Sol. As we know there exist 2 black queens and 2 kings in
a set of 52 playing cards.
2C
1
= 8√2 𝑚.
Total surface area of the sphere = 4𝜋 × 8√2 × 8√2
=512𝜋 m2
S340. Ans.(d)
Sol.
Let number of employees in ‘Adda 247’ initially = n
ATQ –
= (5 + 1)
5n + 26 = 6n + 6
n = 20
New number of employees in ‘Adda 247’ = 20 + 1= 21
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S344. Ans.(b)
Sol. ATQ, vowels have to come together so A and I
together will be treated as a single letter.
And, A and I can change their respective places in 2! Ways.
So, Number of ways = (8-1)! × 2!= 7!× 2!
= 10080 ways
2C
1
85
S343. Ans.(c)
45×45×45
Sol. Number of cubes =
= 216
1
So, Required Probability = 52C1 + 52C1 = 13
Now, radius of the Sphere = 8 × 64√2
(𝑛+1)
Radius of cylinder=r=6cm
Height of cylinder=h=12cm
Volume of cylinder = πr²h
= 432π cm³
1
S339. Ans.(a)
Sol.
Let each of base and height of the isosceles right-angle
triangle is a meter
so its hypotenuse will be a√2 m.
Area of isosceles right-angle triangle = 128 × 16
(5𝑛+26)
Volume of cylinder = πr²h (r → radius of cylinder, h →
height of cylinder)
R = r (given)
ATQ,
4
𝜋𝑅3 =288𝜋 ⇒ R3 =216 ⇒ R=6cm=r
3
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S346. Ans.(d)
Sol.
Let the radius of cylinder and hemisphere be r cm.
So, height of cylinder = 2r cm.
Surface area of cylinder = 2πrh
= 4πr2
Total Surface Area of Hemi-Sphere = 3πr2
Required result =
1
4π𝑟2 −3π𝑟2
3π𝑟2
× 100
=33 3 %
S347. Ans.(e)
Sol.
Possible cases of balls will be 2 red or 2 Orange or 2 Green.
4C
3C
2C
6
3
1
5
Required probability = 9C2 + 9C2 + 9C2 = 36 + 36 + 36 = 18
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S348. Ans.(a)
Sol. In the word BLASTING, there are two vowels (A, I) and
six consonants (B, L, S, T, N, G).
So, required probability =
7!×2!
8!
2
1
=8=4
S349. Ans.(c)
Sol. radius = r cm
Height =3r cm
ATQ
2πr(r+h)=1232
⟹
2×
22
7
S356.Ans.(c)
Sol.
Wrong number = 10
Pattern of series –
8 × 0.5 = 4
4×1=4
4 × 1.5 = 𝟔
𝟔 × 2 = 12
12 × 2.5 = 30
30 × 3 = 90
× 𝑟 × 4𝑟 = 1232
⟹
r = 7cm
Height = h=21cm
22
Volume of cylinder = 7 × 7 × 7 × 21 = 3234 𝑐𝑚3
S350. Ans.(e)
Sol.
Let us suppose number of green balls in the box = x
ATQ,
6C
1
(6+5+x)C
6
𝑥+11
1
1
=3
=
1
3
x+11 =18
∴x=7
S351.Ans.(a)
Sol.
Number of complaints received Tuesday= 100 + 80 +
70 + 110 =360
Number of complaints received on Wednesday= 50 +
60 + 120 + 90 = 320
Required difference= 360 − 320
= 40
S352.Ans.(b)
Sol.
Required %=
(70+110)−(50+60)
(50+60)
70
= 110 × 100 = 63.63
S354.Ans.(c)
Sol.
(70+120)
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 % = (100+80+70+110) × 100
190
= 360 × 100 = 52.77 ≈ 53 %
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S357. Ans.(c)
Sol.
Wrong number = 11
Pattern of series –
𝟏𝟐 + 22 = 16
16 + 32 = 25
25 + 42 = 41
41 + 52 = 66
66 + 62 = 102
102 + 72 = 151
S358. Ans.(d)
Sol.
Wrong number = 25
Pattern of series –
21 + 23 = 𝟐𝟗
𝟐𝟗 − 32 = 20
20 + 23 = 28
28 − 32 = 19
19 + 23 = 27
27 − 32 = 18
S359.Ans.(a)
Sol.
Wrong number = 104
Pattern of series –
20 + 8 = 28
28 + 12 = 40
40 + 16 = 56
56 + 20 = 76
76 + 24 = 𝟏𝟎𝟎
𝟏𝟎𝟎 + 28 = 128
S353.Ans.(c)
Sol.
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑎𝑡𝑖𝑜 = (80 + 60): (50 + 90)
= 1: 1
86
S355. Ans.(d)
Sol.
Required ratio=(100 + 80 + 110): (50 + 60 + 120 + 90)
= 290: 320
= 29: 32
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S360.Ans.(e)
Sol.
Wrong number = 20
Pattern of series –
1×1+1= 2
2×2+2= 6
6 × 3 + 3 = 𝟐𝟏
𝟐𝟏 × 4 + 4 = 88
88 × 5 + 5 = 445
445 × 6 + 6 = 2676
S368. Ans.(a)
Sol.
4? +
4? = 256
?=4
S369. Ans.(e)
Sol.
1512
?
1512
?
?
100
?
100
?=
288 ×100
640
(80x−70x)
80x
× 100
S373. Ans.(b)
Sol.
Let total production in any of these years be 100x
62.5
? = 400
S366. Ans.(b)
Sol.
30+60+30
% of 100x
3
Required ratio = (30+30)%
= 2: 3
of 100x
× 450 + ?2 = 256 – 4
S374. Ans.(d)
Sol.
Let total production in any of these years be 100x
ATQ,
10% of 100x = 12000
x = 1200
10% of 1,20,000 + 30% of 1,20,000
Required average =
2
= 24000
?2 = 252 – 108
? = 12
S367. Ans.(d)
Sol.
× 750 + 110) =
88
100
× 2500
? × 440 = 2200
?=5
87
× 640 = 400 – 112
S372. Ans.(e)
Sol.
Required difference = 60% of 1,50,000 – (20+30) % of
1,50,000
= 15000
250×100
44
40
× 640 + 100 × 280 = 400
= 12.5 %
×? −25 = 225
100
× 400
= 280 – 244
∴ Required percent =
S365.Ans.(d)
Sol.
?×(
70
100
S371. Ans.(d)
Sol.
Let total production in any of these years be 100x
S364.Ans.(e)
Sol.
3167 − 2881 − 121 =? −41
? = 206
24
× 488 =
? = 45
S363.Ans.(c)
Sol.
75 − 63 + 25 =?
? = 37
100
50
100
S370.Ans(d)
Sol.
43−16 = √?− 12
? = 1521
?=
+
? = 42
S362.Ans.(e)
Sol.
100
× 980 = 1040
4? + 784 = 1040
S361.Ans.(a)
Sol.
63 + 18 =?2
?= 9
62.5
80
100
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S375. Ans.(b)
Sol.
Let total production in any of these years be 100x
20
∴
× 100x = 18000
100
x=900
Total production in 2019 =
= 1,08,000.
120
100
S384. Ans.(d)
Sol.
Sum of selling price of all 3 type of refrigerators in year
2016 = (16000+22000+26000) = 64000
Sum of selling price of all 3 type of refrigerators in year
2017 = (14000+25000+32000) = 71000
Sum of selling price of all 3 type of refrigerators in year
2018 = (15000+19000+29000) = 63000
Sum of selling price of all 3 type of refrigerators in year
2019 = (17000+22000+28000) = 67000
So, in Year 2018 it is lowest.
× 100 × 900
S376. Ans.(b)
Sol.
–26, +52, –78, +104, –130
So, 640 – 130 = 510
S385. Ans.(b)
Sol.
15000
Required % = 17000 × 100 = 88.23% ≈ 88% (approx.)
S377. Ans.(e)
Sol. Pattern is —
+25, +50, +100, +200, +400
So, 402+400=802
S386. Ans.(d)
Sol.
I . x2 − 6𝑥 − 8𝑥 + 48 = 0
x(x −6) − 8 (𝑥 − 6) = 0
(x – 8) (𝑥 − 6) = 0
x = 6, 8
II . y2 – 9𝑦 − 8𝑦 + 72 = 0
y(y −9) − 8 (𝑦 − 9) = 0
(y −9) (𝑦 − 8) = 0
y = 9, 8
x ≤y
S378. Ans.(a)
Sol.
Pattern is —
+(5²–1), +(7²+1), +(9²–1), +(11²+1), +(13²–1)
So,
293+(13²–1)=461
S379. Ans.(d)
Sol.
Pattern is —
÷5, ×6, ÷5, ×6, ÷5
So, 50.4 ÷ 5 = 10.08
S380. Ans.(b)
Sol.
×2.5, ×1.5, ×2.5, ×1.5, ×2.5
So, 337.5 × 2.5 = 843.75
S381. Ans.(c)
Sol.
19000
Marked price of C in 2018 = 100−24 × 100 = 25000.
Cost price of C in 2018 =
25000
5
× 3 = 15000
Required Difference= 6000 − 4000 = 2000
S382. Ans.(d)
Sol.
Required average =
16000+25000+15000+22000
4
= 19500 𝑅𝑠.
S383. Ans.(c)
Sol.
19000
Required ratio = 25000 = 19: 25
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S387. Ans.(b)
Sol.
I . x2 + 7x + 6x + 42 = 0
𝑥 (𝑥 + 7) + 6(𝑥 + 7) = 0
(𝑥 + 6)(𝑥 + 7) = 0
x = −6, −7
II. y2 + 8y + 7y + 56 = 0
y(y + 8) + 7(y + 8) = 0
(y + 8) (y + 7) = 0
y = −8, −7
x≥y
S388. Ans.(c)
Sol.
I . x2 + 6x + 2x+ 12 = 0
x(x + 6) + 2(x + 6) = 0
(x + 2) (x + 6) = 0
x = −2, −6
II. 6y2 + 9y + 4y + 6 = 0
3y(2y + 3) + 2(2y + 3) = 0
(2y + 3) (3y + 2) = 0
3
2
y =−2,−3
x<y
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S389. Ans.(a)
Sol.
I . 2x2 + 6x + 3x + 9 = 0
2x(x + 3) + 3(x + 3) = 0
(x + 3) (2x + 3) = 0
3
x = −3, − 2
S395. Ans.(d)
Sol.
Pattern is,
II . y2 + 16y + 12y + 192 = 0
y(y + 16) + 12(y + 16) = 0
(y + 16) (y + 12) = 0
y = −16, −12
x>y
S396. Ans.(e)
Sol. Price of a one kg sugar = 84 ×
S390. Ans.(a)
Sol.
I. x² – 9x + 20 = 0
x² – 5x – 4x + 20 = 0
x (x – 5) –4 (x – 5) = 0
(x – 4) (x – 5) = 0
x = 4, 5
II. 𝑦 2 + 3𝑦 + 3𝑦 + 9 = 0
𝑦(𝑦 + 3) + 3(𝑦 + 3) = 0
(𝑦 + 3)(𝑦 + 3) = 0
𝑦 = −3, −3
𝑥>𝑦
Price of one kg of salt = 840 ×
10
21
11
21
= 𝑅𝑠 44
= 𝑅𝑠40
Required difference = (20 × 44 – 15 × 40)
= 880 – 600
= 𝑅𝑠. 280
S397. Ans.(a)
Sol. Price of one kg of tea =
900
18
= 𝑅𝑠50
1500
Price of one kg of rice = 30 = 𝑅𝑠50
Required % =
50−50
50
× 100 = 0%
S398. Ans.(d)
S391. Ans.(b)
Sol.
63×12
18
Sol. Required ratio = 42×25 = 25
S399. Ans.(b)
Required%=
S392. Ans.(d)
Sol.
20+15
30+12
1
× 100 = 83 %
3
S400. Ans.(b)
Sol. Required sum = (56 × 15) + (32 × 30) + (40 × 25)
= 2800 Rs.
S393. Ans.(e)
Sol.
S394. Ans.(a)
Sol.
Pattern is,
89
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